In this article, we shall study the factors affecting the rate of a chemical reaction and the Arrhenius equation.

Factors Affecting the Rate Of Reaction:

The Concentration of Reactants:

The number of collisions and hence the activated collisions between the reactant molecules increase with the increase in concentration. According to the collision theory, the rate of a reaction should increase with the increase in the concentration since the rate is directly proportional to the collision frequency.

Pressure:

The number of collisions increases with an increase in the partial pressures of gases. Hence the rate of a reaction involving gaseous reactants increases with an increase in partial pressures. However, it has no effect on reactions involving reactants in liquid or solid phases.

It is important to keep in mind that the partial pressures of reactants can be increased by increasing the pressure of the overall system. However, the partial pressures do not increase when an inert gas or a non-reacting gas is added to the reaction mixture at constant volume.

Temperature:

The average kinetic energy increases with increase in absolute temperature. Hence the number of molecules with energy greater than the threshold energy also increases.

As a result, the number of effective collisions between reactant molecules also increases. Therefore, usually, it is observed that the rate of reaction increases with increase in temperature.

Catalyst:

The catalyst is a substance which changes the rate of a reaction without being consumed or without undergoing any chemical change during the reaction.

In case of reversible reactions, the catalyst lowers the activation energies of both forward and backward reactions to the same extent and helps in attaining the equilibrium quickly.

Some substances may decrease the rate of a reaction. These are generally referred to as negative catalysts or inhibitors. They interfere with the reaction by forming relatively stable complexes, which require more energy to break up. Thus the speed of the reaction is reduced.

Nature of Reactants:

The rate of a reaction depends on the nature of bonding in the reactants. Usually, the ionic compounds react faster than covalent compounds.

The reactions between ionic compounds in water occur very fast as they involve the only exchange of ions, which were already separated in aqueous solutions during their dissolution.

Whereas, the reactions between covalent compounds take place slowly because they require energy for the cleavage of existing bonds.

The Orientation of Reacting Species:

The reaction between the reactants occurs only when they collide in the correct orientation in space. Greater the probability of collisions between the reactants with proper orientation, greater is the rate of reaction.

The orientation of molecules affects the probability factor, p. The simple molecules have more ways of proper orientations to collide. Hence their probability factor is higher than that of complex molecules.

The orientation factor also affects the interaction between reactants and catalysts. For example in case of biological reactions, which are catalyzed by enzymes, the biocatalysts. The enzymes activate the reactant molecules (or substrates) at a particular site on them. These sites are called active sites and have definite shape and size. The size, stereochemistry, and orientation of substrates must be such that they can fit into the active site of the enzyme. Then only the reaction will proceed. This is also known as lock and key mechanism. The enzymes lose their activity upon heating or changing the pH or adding certain chemical reagents. This is due to deformation of the configuration of the active site.

Surface Area of Reactant:

The rate of a reaction increases with increase in the surface area of solid reactant if any used. The surface of a solid can be increased by grinding it to a fine powder.

This is also true with the solid catalysts, which are usually employed in finely powdered form while carrying out a chemical reaction.

The Intensity of Light:

The rate of some photochemical reactions, which occur in presence of light, increases with increase in the intensity of suitable light used.

With the increase in the intensity, the number of photons in light also increases. Hence number of reactant molecules get energy by absorbing more number of photons and undergo a chemical change.

Nature of Solvent:

The solvent may affect the rate in many ways as explained below:

The solvents are used to dissolve the reactants and while doing so they help in providing more interactive surface between reactant molecules which may be otherwise in different phases or strongly bonded in the solid phase.

Usually, solvents help in breaking the cohesive forces between ions or molecules in the solid state. The polar molecules tend to dissolve more in polar solvents with more dielectric constants and react faster in them. Whereas nonpolar molecules prefer nonpolar solvents.

In case of diffusion controlled reactions, the viscosity of the solvent plays a major role. The rate decreases with increase in the viscosity of the solvent.

Effect of Change of Temperature on the Rate of Reaction:

The average kinetic energy increases with increase in absolute temperature. Hence the number of molecules with energy greater than the threshold energy also increases.

As a result, the number of effective collisions between reactant molecules also increases. Therefore, usually, it is observed that the rate of reaction increases with increase in temperature.

The two distribution graphs are shown below for a lower temperature T₁ and a higher temperature T₂. The area under each curve represents the total number of molecules whose energies fall within the particular range.

The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of E_a that are shown.

It is clear from these graphs that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) proceed more rapidly at higher temperatures.

Arrhenius Equation:

Arrhenius came up with an equation that demonstrated that rate constants of different kinds of chemical reactions varied with temperature. This equation indicates a rate constant that has a proportional relationship with temperature. For example, as the rate constant increases, the temperature of the chemical reaction generally also increases. The result is given below:

Where k is rate constant, Ea the activation energy, T the absolute temperature of the reaction and R is universal gas constant. A is called frequency factor or pre-exponential factor and proportional to the frequency of collisions between reacting molecules. A is independent of the absolute temperature T.

Equations (1), (2) and (3) are different forms of Arrhenius equation. A and Ea are called Arrhenius parameters.

Arrhenius Equation and Temperature Variation:

The relation between rate constant k and the absolute temperature T of the reaction is given by

For two different temperatures say T₁ and T₂ we have

Determination of Activation Energy:

We have

By knowing Values of K₁ and K₂ at temperatures T₁ and T₂ using experiments, the value of activation energy can be calculated.

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The occurrence of a bimolecular chemical reaction can be explained on the basis of collision theory.

The Collision of Reacting Molecules:

Consider a bimolecular general reaction

A  + B  →   C

In order for a chemical reaction to take place, the molecules of reactants A and B must collide. The rate of a chemical reaction depends on the rate of collision between the molecules. As the concentration and temperature increase, the rate of reaction also increases.

The Energy of Activation:

The collision between the molecules in a chemical reaction provides the kinetic energy needed to break the necessary bonds so that new bonds can be formed.  If the kinetic energy is not sufficient the bond between the reactants will not be broken and thus new bonds will not be formed. This required energy is called activation energy.

The activation energy (E_act) is defined as the minimum kinetic energy required for the molecular collision to lead to the reaction.

Thus reaction to take place the kinetic energy of colliding molecules should be greater than the activation energy.

The Orientation of Reacting Molecules:

The minimum kinetic energy (energy of activation) does not mean successful collision leading to the reaction.

For successful collision leading to the reaction, the colliding molecules must be so oriented relative to each other that the group reacting or bonds to be shifted are relatively close. This criterion is not essential for simple molecules but it becomes very much essential for complex molecules.

Potential Energy Barrier:

Consider a bimolecular general reaction

B  + AC   →   BA   +    C

During the collision the electron distribution about the three nuclei A, B and C changes in such a way that the new bond B-A strengthens at the same time the old bond A-C weakens. A stage is reached when all the three nuclei are weakly linked together. This state is called activated complex or transition state.

B   +   AC   →     B—-A —-C

The transition state is an unstable transitory complex of the highest energy state through which the reactants must pass on the way to products.  It decomposes spontaneously to form products.

To achieve this configuration, atoms require energy to overcome the repulsion between B and AC that have filled shells of electrons. This energy comes from kinetic energy due to the collision of reacting molecules and gets converted into potential energy in the activated complex. Thus transition state has the highest energy state. In an energy profile diagram (E.P.D.), the highest energy point (Peak) corresponds to the transition state.

Thus reactant molecules have to climb up the barrier to get converted into products.

Mathematical Treatment to Collision Theory:

Consider a bimolecular general reaction

B  + AC    →    BA   +    C

It is a second-order reaction. Hence the rate of collision is given by

Rate of collision = Z [AC] [B]

Where Z = frequency of collision.

The reaction rate is given by

Rate of reaction = P x f x Rate of collision

Where P is the fraction of collisions with proper orientations of colliding molecules. f is a fraction of molecules with sufficient kinetic energy.

Rate of reaction = P x f x Z [AC] [B]  …… (1)

Burt the rate of second-order reaction is given by

Rate of reaction = k [AC] [B] …… (2)

From (1) and (2) we have

k = P.f.Z

By Arrhenius equation we have

Where A = P.Z and called as frequency factor or pre-exponential factor.

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In this article, we shall study the molecularity of reaction and catalysis.

The Concept of Elementary Reactions:

Many reactions that follow a simple rate law are actually taking place in series of steps. These reactions are called complex reactions. Each step in a complex reaction is called elementary reaction. Thus complex reaction can be broken down into the elementary reactions.

An elementary reaction is defined as a reaction that takes place in a single step and can’t be broken down further into simplest chemical reactions.

Illustration:

Consider complex reaction

3 ClO^–_(aq)  →  ClO₃^–_(aq) + 2Cl^–_(aq)

Actually, this reaction takes place in two steps. Thus there are two elementary reactions.

Step-1:

2 ClO^–_(aq)  →  ClO₂^–_(aq) +  Cl^–_(aq)    (Bimolecular reaction)

Step – 2:

ClO₂^–_(aq)  +  ClO^–_(aq)  →  ClO₃^–_(aq) +  Cl^–_(aq)    (Bimolecular reaction)

The sum of the two reactions gives the overall reaction.

Molecularity of Elementary Reactions:

The molecularity of an elementary reaction is defined as the number of reaction molecules taking part in the reaction.

Example of Unimolecular Reaction:

O_3(g)  →  O_2(g) +  O_(g) 

C₂H₅I_(g)  →  C₂H_4(g) +  HI_(g) 

Example of Bimolecular Reaction:

O_3(g)  +  O_(g)  →  2O_2(g)

2NO_2(g)  → 2NO_(g)   +  O_2(g)

Distinguishing Between Molecularity and Order of Reaction:

Molecularity of a Reaction:

• The molecularity of a reaction is defined as the number of reaction molecules taking part in the reaction.
• Molecularity is always a whole number.
• It is a theoretical property indicating the number of molecules involved in each act leading to the reaction.
• It does not change with experimental conditions.
• It is the property of elementary reaction and has no meaning for a complex reaction.

Order of a Reaction:

• The overall order of the reaction is defined as the sum of the exponents to which the concentration terms in the rate law are raised.
• order of reaction may be an integer, fraction, or zero.
• It is a purely experimental property indicating the dependence of the observed reaction rate on the concentration of the reactants.
• It may change with experimental conditions.
• It is the property of both elementary and complex reactions.

Multistep Reactions:

A multistep reaction is a reaction involving two or more steps. Consider the reaction

A       →     C

It consists of two steps.

First step:         A    →    B

Second step:    B    →   C

In the above reaction, B is intermediate. The Intermediate is the product of the first step and reactant of the second step.

Reaction Intermediates:

The additional species other than the reactants or products formed in the mechanism during the progress of a reaction is called reaction intermediate.

Characteristics of Intermediate:

• They may be stable or unstable.
• The number of Intermediates in a reaction =  The number of Steps  –   1
• Thus reaction involving two steps will have one intermediate and one step reaction will have no intermediate.
• The intermediates appear in mechanism but do not appear in overall reaction because they are produced in one step and consumed in another step.
• The concentration of reaction intermediates is very small, hence cannot be determined easily.
• The rate of reaction is independent of the concentration of intermediates.
• The life period of reaction intermediates is very small hence they cannot be isolated.

Catalysis:

Catalyst and its Effect on the Rate of Reaction:

A catalyst is a substance, added to the reactants, that increases the rate of reaction without itself being consumed in the reaction

Example: In preparation of O₂ from KClO₃ in laboratories MnO₂ is used as a catalyst.

Characteristics of Catalyst:

• The catalyst does not appear in an overall reaction because they are consumed in one step and regenerated in another step.
• A catalyst lowers the activation energy of a reaction.
• In presence of a catalyst the height of the energy barrier decreases. Thus the number of molecules the possess the minimum kinetic energy increases.
• Chemically, the catalyst remains unchanged during a reaction.
• Catalyst does not change the quantity of the product.
• A catalyst is specific, which means different chemical reactions may have a different catalyst.
• Just a small amount needed to achieve a big increase in the rate of reaction.
• More amount of catalyst used can further increase the rate of reaction.
• A catalyst in powder form can further increase the rate of reaction.
• A catalyst may undergo a physical change in a reaction.

Distinguishing Between Catalyst and Reaction Intermediate:

Catalyst:

• A catalyst is a substance, added to the reactants, that increases the rate of the reaction without itself being consumed in the reaction.
• A catalyst increases the rate of a reaction.
• A catalyst is present at the start of the reaction.
• A catalyst is consumed in one step and regenerated in the subsequent step.
• The concentration of catalyst may appear in rate law.
• Catalysts are stable under ordinary conditions.

Intermediate:

• The additional species other than the reactants or products formed in the mechanism during the progress of a reaction is called reaction intermediate.
• Intermediate has no effect on the rate of reaction.
• An intermediate exist during the mechanism of the reaction.
• An intermediate is produced in one step and consumed in the subsequent step.
• The concentration of intermediate does not appear in rate law.
• Intermediates are highly unstable and ha a short life.

Note:

Both catalyst and intermediate do not appear in overall reaction.

Rate Determining Step (R.D.S.)

In a multistep reaction, rate of overall reaction depends upon the rate of the slowest step.  This slowest step is called rate determining step.

Example : Consider substitution reaction

R-X       +      Y        →      R-Y     +      X

Substrate   Reagent       Product    Living group

This reaction takes place as  follows

First Step:

R – X   →   R⁺  +     : X^–

Second Step:

R⁺  +   : Y^–  →   R  Y

In above case rate of overall reaction depends on the rate of the first step. Hence the first step is labelled as R.D.S.

The overall reaction cannot take place faster than the rate of rate determining step. Hence RDS step determines rate of overall reaction. As RDS is elementary reaction, the rate law can be determined from its stoichiometric equation. In rate law the exponents are equal to the coefficient of balanced equation for the step.

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In this article, we shall study the analytical treatment to the zero-order reaction, and the rate of zero-order reaction.

Order of Reaction:

The overall order of the reaction is defined as the sum of the exponents to which the concentration terms in the rate law are raised.

Let us consider a general reaction

aA  + bB  → Products

The rate law can be written as

Rate = K  [A]^x [B]^y …………… (1)

Thus the overall order of the reaction is (x + y).

Zero Order Reaction:

Integrated Law for Zero-Order Reaction:

A reaction whose rate is independent of the concentration of reactants is called a zero-order reaction.

Let us consider a general reaction

Let [A]_o be initial concentration of A (i.e at t = 0)  and be final concentration of A (i.e at t = t)

Integrating both sides of above equation

This relation is known as integrated law for zero-order reaction.

Expression for the Integrated Rate Constant:

From equation (1) we have

This is an expression of the integrated rate constant for the zero-order reaction.

Unit of Integrated Rate Constant:

The unit of  integrated rate constant is  mol dm^-3 t^-1 ( mol dm^-3 s^-1 or  mol dm^-3 min^-1,  mol dm^-3hr^-1)

Half-Life of Zero Order Reaction:

The half-life of a reaction is defined as the time required for the reactant concentration to fall to one half of its initial value. Thus for t = t_1/2,   [A]_t  = ½[A]_o

The  integrated rate constant for the zero-order reaction is given by

This is an expression of the half-life of a zero-order reaction.

Graphical Representation of Zero Order Reaction in Different Ways:

The graph of rate of reaction against time:     

The differential rate law for the zero-order reaction is

Thus the graph of the rate of reaction versus time is a straight line parallel to the time axis.

The graph of Rate of reaction against Initial Concentration:  

The differential rate law for the zero-order reaction is

Thus the rate is independent of the concentration of reactants. Hence Thus the graph of the rate of reaction versus concentration of reactants is a straight line parallel to concentration axis.

The graph of Concentration of the reactants against time:

The differential rate law for the zero order reaction is

[A]_t  = – kt + [A]_o

It is of the form y = mx + c. Thus the graph of concentration at instant versus time is a straight line with y-intercept. [A]_o. The slope of the straight line is – k.

Examples of Zero-order Reaction

• Decomposition of NH₃ on a hot platinum surface:

• Decomposition of Nitrous oxide in presence of platinum as a catalyst:

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In this article, we shall study the order of reaction and the analytical treatment to the first-order reaction, and the rate of the first-order reaction.

Order of Reaction:

The overall order of the reaction is defined as the sum of the exponents to which the concentration terms in the rate law are raised.

Let us consider a general reaction

aA  + bB  → Products

The rate law can be written as

Rate = K  [A]^x [B]^y …………… (1)

Thus the overall order of reaction is (x + y).

Example:

In the reaction, 

NO_2(g)   +   CO_(g)  →  NO_(g)   + CO_2(g)

The rate of reaction is experimentally found to proportional

to [NO₂]² and independent of [CO]. Thus x = 2 and y = 0

Thus the rate law of reaction is

Thus the overall order of the reaction is 2.

Characteristics of Order of Reaction:

• Order of reaction represents the number of atoms, ions and molecules whose concentration influence the rate of reaction.
• Order of the reaction is defined as the sum of the exponents to which the concentration terms in the rate law are raised. Thus it is not dependent on the stoichiometric coefficients in a balanced chemical reaction.
• Values of x and y are determined experimentally. The values of x and y in the rate law are not necessarily equal to the stoichiometric coefficients of reactants. Thus the order of reaction can be decided by performing experiment only.
• Order of reaction is defined in terms of concentration of reactants only and not of products.
• Order of reaction may be integer, fraction or zero.

Reactions with No Order:

If the rate of reaction cannot be expressed in the form, Rate, then the reaction has no order and term order should not be used for such reactions.

Example:

Clear Concept of Stoichiometric Coefficients of Balanced Chemical Equations and Exponents in Rate Law:

Let us consider a general reaction

aA  + bB  → Products

The stoichiometric coefficients of A and B are a and b.

Let us assume that the rate of reaction depends on [A]^x and [B]^y

The rate law is written as

Rate = K  [A]^x [B]^y …………… (1)

The values of x and y for the reaction are found experimentally. The values of x and y may be integer, fraction or zero.

First Order Reactions:

Integrated Rate Law For First Order Reaction:

The equations which are obtained by integrating the differential rate laws and which gives the direct relationship between the concentrations of the reactants and time is called integrated rate laws. A reaction whose rate depends on the single reactant concentration is called the first-order reaction.

Let us consider a general reaction

Let [A]_o be initial concentration of A (i.e at t = 0)  and [A]_t be the final concentration of A (i.e at t = t)

Integrating both sides of above equation

This relation is known as exponential integrated law for first order reaction.

The Expression for the Integrated Rate Constant for First Order Reaction:

This is an expression for the integrated rate constant for the first order reaction.

The expression for the integrated rate constant in terms of initial concentration for First Order Reaction:

Let a mol dm^-3 be the initial concentration of A (i.e at t = 0) and at some instant ‘t’ the decrease in concentration is x mol dm^-3.  (i.e at t = t). Thus   [A]_o = a and  [A] = a – x

Substituting in equation (2) we have

This is an expression for the integrated rate constant for the first order reaction in terms of initial concentration.

Unit of  the Integrated Rate Constant for the First Order Reaction:

The integrated rate constant for the first order reaction is given by

The quantity [A]_o / [A]  is a pure ratio. Hence it has no unit. Thus the unit of  integrated rate constant is per unit time (s^-1 or  min^-1 , hr^-1 )

Half-Life of First Order Reaction:

The half-life of a reaction is defined as the time required for the reactant concentration to fall to one half of its initial value. Thus for t = t1/2,  [A] = ½ [A]_o

The  integrated rate constant for the first order reaction is given by

This is an expression for the half-life of the first-order reaction.

Graphical Representation of Half-Life:

A graph is drawn by plotting time as a multiple of half-life on the x-axis and the concentration of reactant in terms of original concentration on the y-axis at that instant. The graph is as follows.

The graph shows that it is an exponential process. Thus this process will never complete. i.e. the graph will never touch x-axis.

Note: Such graph is shown by the disintegration of a radioactive element. The radioactive elements obey decay law.

Graphical Representation of the First Order Reaction in Different Ways:

The graph of Rate of reaction against time:     

The differential rate law for the first order reaction is

It is of the form y = mx + c. Thus the graph of rate reaction versus concentration at an instant  is a straight line passing through the origin (since c = 0). The slope of the straight line is k.

The graph of Concentration of the reactants of against time:

The exponential rate law for the first order reaction is [A]_t = [A]_o e^-kt

Thus it is an exponential process. Thus this process will never complete. i.e. the graph will never touch x-axis.

The graph of Concentration of the reactants against time:

This equation is of form y = mx + c. Thus the graph of  log₁₀[A]_t  versus time is a straight line with y-intercept log₁₀[A]₀.

The graph of log₁₀([A]₀/ [A]_t) against time:

This equation is of form y = mx + c. Where c = 0. Thus the graph of  log₁₀([A]₀/ [A]_t) versus time is  a straight line passing through the origin

Examples of First-order Reactions:

Pseudo First Order Reaction:

The reactions that have higher order true rate law are found to behave as the first order are called pseudo-first order reactions.

Examples of Pseudo First Order Reactions:

• Hydrolysis of methyl acetate:

CH₃COOCH_3(aq)  + H₂O_(l) → CH₃COOH_(aq)) + CH₃OH_(aq)

The true rate law of reaction must be

Rate = K'[CH₃COOCH₃][H₂O]

Thus it seems that the rate of reaction is dependent on two reactants.

But the concentration [H₂O]  is constant (say k’’)

Rate = K’ K”[CH₃COOCH₃]

Hence,  Rate = K[CH₃COOCH₃]

Thus the reaction actually a first order reaction. Hence it is called as pseudo first order reaction.

• Hydrolysis of cane sugar (sucrose):

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In this article, we shall discuss the concept of chemical kinetics, rate of reactions, and types of reactions on the basis of their rates.

Rate of Reaction:

The branch of chemistry, which deals with the rate of chemical reactions, the factors affecting the rate of reactions and the mechanism of the reaction. is called chemical kinetics.

Importance of Chemical Kinetics:

• It gives an idea that how fast a reaction can occur. using which we know how quickly a medicine is able to work or what is the time required for completion of the reaction.
• In environmental chemistry, it is important to study the ozone balance in the upper atmosphere. The maintenance and depletion of the ozone layer depend on the relative rate of formation and destruction of ozone in the upper atmosphere.
• It is important in catalysis. It is used to solve industrial problems such as the development of catalysts to synthesize new materials.
• It has biological importance. In our body, large protein molecules called enzymes, increase the rate of biological reactions.
• It is important in the food industry. It is used to determine the factors which spoil the food and the rate at which the food is getting spoilt. Hence the probable expiry date or best before the date can be determined.
• The fast setting ceramic material is used for the material for dental fillings.
• It is used to determine the rate at which steel rusts.
• It is used to study the rate at which fuel burns in an automobile engine.

Classification of Chemical Reactions on the Basis of the Rate of the Reaction:

Fast/Instantaneous Reactions (Type – I):

The chemical reaction which completes in less than 1ps (one pieco second) (10^-12 s) time, is known as the fast reaction. It is practically impossible to measure the speed of such reactions. The reason for a very fast rate of such reaction is that no chemical bonds are to be broken among the reactants.

e.g., ionic reactions. Organic substitution reactions, neutralization reaction.

Very Slow Reactions (Type – II):

Chemical reactions which complete in a long time from some minutes to some years are called slow reactions. The rates of such reactions are hardly of any physical importance.

e.g. rusting of iron, transformation of carbon into diamond etc.

Moderately Slow Reactions:

Chemical reactions which are intermediate between slow and fast reactions are called moderately slow reactions.

These reactions proceed at a moderate speed which can be measured easily. Mostly these reactions are in molecular nature.

Note:

The rate of chemical reaction can be changed by changing the conditions under which they occur.

Rate of a Reaction:

The rate of a chemical reaction may be defined as the change in concentration of designated species (reactant or product) per unit time. The rate can be measured qualitatively or quantitatively.

Qualitative Rate: The qualitative rate is based on certain visual parameters such as disappearance of reactants, colour change, effervescence etc.

Quantitative Rate: The quantitative rate is based on the rate of decrease in the concentration of any one reactant or the rate of increase in the concentration of any one product.

Average Rate of Reaction:

The average rate of reaction is defined as the change in concentration of reactant or product divided by the time interval over which the change occurs.

Consider a general reaction

Positive sign indicates that the concentration of B is increasing. It is to be noted that the rate of reaction is always positive.

The rate of reaction is expressed in terms of moles per litre per unit time (mol L^-1t^-1) or molar per unit time (Mt^-1). If time is measured in a second then the unit is moles per litre per second (mol L^-1s^-1) or molar per second (Ms^-1). In terms of partial pressure (for gaseous reactions) units is atmosphere per unit time. The average rate of reaction depends on the time interval chosen.

Average Rates of Some Reactions:

Example – 1:

In this case, stoichiometric coefficients of the reactants and products are same, the rate of the reaction is

Example – 2:

In this case, stoichiometric coefficients of H₂, N₂ and NH₃ are  1, 3 and 2 respectively. Hence the rate of the reaction is

Example – 3:

In this case, stoichiometric coefficients of N₂O₅, NO₂ and O₂ are  2, 4 and 1 respectively. Hence the rate of the reaction is

General Example:

aA  + bB  → cC   +   dD

In this case, stoichiometric coefficients of A, B, C and D are a, b, c and d respectively. Hence the rate of the reaction is

Instantaneous of Rate of a Reaction:

The rate of a reaction at a specific instant is called an instantaneous rate.

If the average rate of reaction is calculated for a shorter and shorter interval of time, a rate at a specific instant can be obtained. The instantaneous rate at the beginning of a reaction is called the initial rate of the reaction.

Consider a general reaction

Example – 1:

In this case, stoichiometric coefficients of H₂, N₂ and NH₃ are  1, 3 and 2 respectively. Hence the rate of the reaction is

General Example:

aA  + bB  → cC   +   dD

In this case, stoichiometric coefficients of A, B, C and D are a, b, c and d respectively. Hence the rate of the reaction is

Note:

In chemical kinetics, we deal only with the instantaneous rate of reaction only. Hence the instantaneous rate of reaction is referred as the rate of reaction only.

Graphical Representation of Instantaneous and Average Rate of Reaction in terms of Reactants:

Graphical Representation of Instantaneous and Average Rate of Reaction in terms of Products:

Determination of Rate of Reaction:

Determination of Average Rate of Reaction:

A graph is drawn by taking a concentration of species (reactant or product) on y-axis and time on the x-axis.

The average rate of reaction at time t can be obtained by the change in concentration (C₂ – C₁) of species (reactant or product) in the time interval t1 and t2. (t1 and t2 are equidistant from t) Then the average rate of reaction is calculated using following formula.

Determination of Instantaneous Rate of Reaction:

A graph is drawn by taking the concentration of species (reactant or product) on y-axis and time on the x-axis.

The instantaneous rate of reaction at time t can be obtained by drawing a tangent at time t and finding its slope at that point. The slope of the tangent to the curve at that point gives an instantaneous rate of reaction.

Reaction Life Time:

It is defined as the time taken by a reaction to proceed to 98% of completion. Shorter the life time, faster is the reaction. It is used to compare the rate of reactions.

Rate Laws and Rate Constant:

Law of Mass Action:

This law was given by Goldberg and Waage in 1864. It states that “at a given temperature, the rate of reaction at a particular instant is proportional to the product of the active masses of the reactants at that instant raised to powers which are numerically equal to the numbers of their respective molecules in the stoichiometric equation describing the reaction.”

Explanation:

Consider general reaction

aA  + bB  → cC  +  dD

By law of mass action

Rate ∝  [A]^a [B]^b

∴   Rate = K  [A]^a [B]^b

Rate Law:

The rate of chemical reaction is found to be proportional to the molar concentrations of the reactants raised to simple powers.

Let us consider a general reaction

aA  + bB  → Products

Let us assume the rate of reaction depends on[A]^x and [B]^y.

Thus, Rate ∝  [A]^x [B]^y

Rate = K  [A]^x [B]^y …………… (1)

But instantaneous rate of reaction is given by

This relation is known as the differential rate law or simply rate law. Where k is a constant called specific reaction rate constant or simply rate constant.

Definition:

The rate law is defined as an experimentally determined equation that expresses the rate of a chemical reaction in terms of molar concentrations of the reactants.

Rate Constant:

By rate law we have,  Rate = K  [A]^x [B]^y

If [A] = 1 and [B] = 1, Then

Rate = k = Rate Constant

The rate constant is defined as the rate of reaction would have if all the concentrations were set equal to unity.

Characteristics of Rate Constant:

• The values of the rate constant give an idea about the speed of the reaction. Greater the value of the rate constant, faster is the reaction.
• Each reaction has a definite value of the rate constant at a particular temperature.
• The value of the rate constant depends on temperature.
• The value of rate constant is independent of the concentration of reacting species.
• The value of rate constant depends on nature of reactant, the presence of catalyst, solvent and pH of a solution.
• The unit of rate constant depends on the order of the reaction. In general unit of rate constant is

Where n is the order of reaction.

Notes:

• The value of x and y in the rate law are not necessarily equal to the stoichiometric coefficients (a and b) of  A and B.
• Values of x and y are determined experimentally.

Example to Illustrate Rate Law:

Example:

In the reaction, NO_2(g)   +   CO_(g)  →  NO_(g)   + CO_2(g)

The rate of reaction is experimentally found to proportional

to [NO₂]² and independent of [CO]. Thus x = 2 and y = 0

Thus the rate law of reaction is

Applications of Rate Law:

• The rate law can be used to estimate the rate of a reaction for any given composition of the reaction mixture.
• It can be used to estimate the concentration of reactants and products at any time during the course of reaction.\
• The rate law is useful for prediction of the mechanism of a complex reaction.

Characteristics of Rate of Reaction:

• It is the speed at which the reactants are converted into products at any instant of time.
• It is dependent on the concentrations of the reactant species at that instant.
• As the reaction proceeds the rate of reaction decreases.

Characteristics of Rate Constant:

• It is constant of proportionality in rate law expression.
• It is independent of the concentration of reactants.
• It is constant throughout the reaction i.e. it is independent of the progress of the reaction.

Science > Chemistry > Chemical Kinetics > Introduction to Chemical Kinetics

The post Introduction to Chemical Kinetics appeared first on The Fact Factor.

Science > Chemistry > Solutions and Their Colligative Properties > Short-cut Methods For Calculating Concentration of Solutions

In this article, we shall study short-cut methods to calculate molality, molarity, etc.

These methods can only be used in competitive exams only.

Direct Formulae to Calculate Molality and Molarity:

Where M = molarity in mol L^-1 or M

m = molality in mol kg^-1 or m

a = % by mass of solute

d = density of solution in g/mL or g cm^-3.

M_B = Molecular mass of solute in grams

M_A = Molecular mass of solvent in grams

Note: When using these formulae, take care that the quantities are in prescribed units

Molecular masses of certain substances in grams:

Water H₂O (18), Benzene C₆H₆ (78), Sodium hydroxide NaOH (40), Hydrogen chloride HCl (36.5), Sulphuric acid H₂SO₄ (98), potassium hydroxide KOH (56), Acetic acid (60), Sodium carbonate Na₂CO₃ (116),

Numerical Problems to Calculate Molality and Molarity:

Example – 01:

The density of a solution containing 13 % by mass of sulphuric acid is 1.09 g/mL. Calculate molarity and normality of the solution

Given: a = 13, d = 1.09 g/mL

To Find: Molarity (M) =? and Normality (N) =?

Solution:

n = Molecular mass/equivalent mass = 98 g/49 g = 2

Normality = molarity x n = 1.446 x 2 = 2.892 N

Example – 02:

The density of 2.03 M solution of acetic acid (molecular mass = 60) in water is 1.017 g/mL. Calculate molality of solution

Given: M = 2.03, M_B = 60 g mol^-1, d = 1.017 g/mL

To Find: Molality (m) = ?

Solution:

molality = m = 1/0.4410 = 2.268 molal

Example – 03:

The density of 10.0% by mass of KCl solution in water is 1.06 g/mL. Calculate the molality, molarity and mole fraction of KCl.

Given: a = 10, d = 1.06 g/mL

To Find: Molarity (M) =?, molality (m) =?, mole fraction (X_B) =?

Solution:

Ans: Molarity 1.42 M, Molality = 1.491 m, Mole fraction = 0.0261

Example – 04:

0.8 M solution of H2SO4 has a density of 1.06 g/cm³. calculate molality and mole fraction

Given: M = 0.8 M, d = 1.06 g/cm³.

To Find: Molality (m) =?, mole fraction (X_B) =?

Solution:

molality = m = 1/1.227 = 0.814 molal

0.814 x 18 x (1 – X_B) = 1000 X_B

14.652 – 14.652 X_B = 1000 X_B

1014.652 X_B = 14.652

X_B = 14.652/1014.652  = 0.014

Example – 05:

A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of solution.

Given: M = 6.90 M, a = 30

To Find: density of solution = d = ?

Solution:

Ans: Density of solution = 1.288 g/mL

Example – 06:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm^-3. Calculate molality, molarity and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Given: a = 10, d =  1.070 g cm^-3,

To Find: mole fraction =? molarity = ? and molality =?

Solution:

Example – 07:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given: molality = 2 molal

To Find: Mole fraction =?

Solution:

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Short-cut Methods For Calculating Concentration of Solutions appeared first on The Fact Factor.

In this article, we shall study numerical problems to calculate molality of a solution.

Example – 01:

7.45 g of potassium chloride (KCl) was dissolved in 100 g of water. Calculate the molality of the solution.

Given: mass of solute (KCl) = 7.45 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol^-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 7.45 g/ 74.5 g mol^-1 = 0.1 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1 mol /0.1 kg = 1 mol kg^-1

Ans: The molality of solution is 1 mol kg^-1 or 1 m.

Example – 02:

11.11 g of urea (NH₂CONH₂) was dissolved in 100 g of water. Calculate the molarity and molality of the solution. Given N = 14, H = 1, C = 12, O = 16.

Given: mass of solute (urea) = 11.11 g, mass of solvent (water) = 100 g = 0.1 kg

To Find: Molarity of solution =?

Solution:

Molecular mass of urea (NH₂CONH₂) = 14 g x 2 + 1 g x 4 + 12 g x 1 + 16 g x 1

Molecular mass of urea (NH₂CONH₂) = 60 g mol^-1

Number of moles of solute (urea) = given mass/ molecular mass

Number of moles of solute (urea) = 11.11 g/ 60 g mol^-1 = 0.1852 mol

Volume of water = mass of water/ density = 100 g/1 g mL^-1 = 100 mL = 0.1 L

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.1852 mol /0.1 L = 1.852 mol L^-1 or 1.852 mol dm^-3

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1852 mol /0.1 kg = 1.852 mol kg^-1

Ans: The molarity of solution is 1.852 mol L^-1 and the molality is 1.852 mol kg^-1

Example – 03:

34.2 g of sugar was dissolved in water to produce 214.2 g of sugar syrup. Calculate molality and mole fraction of sugar in the syrup. Given C = 12, H = 1 and O = 16.

Given: Mass of solute (sugar) = 34.2 g, Mass of solution (sugar syrup) = 214.2 g

To Find: Molality and mole fraction =?

Solution:

Mass of Solution = Mass of solute + mass of solvent

Mass of solvent = mass of solution – mass of solute = 214.2 g – 34.2 g = 180 g = 0.180 kg

Molar mass of sugar (C₁₂H₂₂O₁₁) = 12 g x 12 + 1 g x 22 + 16 g x 11 = 342 g mol^-1

Number of moles of solute (sugar) = n_B = Given mass/ molecular mass = 34.2 g/342 g mol^-1  = 0.1 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1 mol /0.180 kg = 0.5556 mol kg^-1

Molar mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Number of moles of solvent (water) = n_A = Given mass/ molecular mass = 180 g/18 g mol^-1  = 10 mol

Total number of moles = n_A + n_B = 0.1 + 10 = 10.1 mol

Mole fraction of solute (sugarl) = x_B = n_B/(n_A + n_B) = 0.1/10.1 = 0.0099

Mole fraction of sugar = 0.0099

Ans: Molality of solution = 0.5556 mol kg^-1 and mole fraction of sugar = 0.0099

Example – 04:

10.0 g KCl is dissolved in 1000 g of water. If the density of the solution is 0.997 g cm^-3, calculate a) molarity and b) molality of the solution. Atomic masses K = 39 g mol^-1, Cl = 35.5 g mol^-1.

Given: the mass of solute (KCl) = 10 g, the mass of solvent (water) = 1000 g = 1 kg, density of solution = 0.997 g cm^-3,

To Find: molarity =? molality = ?

Solution:

Molecular mass of KCl = 39 g x 1 + 35.5 g x 1 = 74.5 g mol^-1

Number of moles of solute (KCl) = given mass/ molecular mass

Number of moles of solute (KCl) = 10 g/ 74.5 g mol^-1 = 0.1342 mol

Molality = Number of moles of solute/Mass of solvent in kg

Molality = 0.1342 mol /1 kg = 0.1342 mol kg^-1

Mass of solution = 10 g + 1000 g = 1010 g

Volume of solution = mass of solution/density = 1010/0.997 g cm^-3

Volume of solution = 1013 cm³ = 1013 mL = 1.013 L

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.1342 mol /1.013 L = 0.1325 mol L^-1

Ans: The molarity of the solution is 0.1325 mol L^-1 or 0.1325 M, the molality of the solution is 0.1342 mol kg^-1 or 0.1342 m.

Example – 05:

Calculate molarity and molality of the sulphuric acid solution of density 1.198 g cm^-3 containing 27 % by mass of sulphuric acid.

Given: density of the solution = 1.198 g cm^-3, % mass of sulphuric acid = 27%,

To Find: Molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of H₂SO₄ = 27 g and mass of H₂O = 100 – 27 g = 73 g = 0.073 kg

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Number of moles of H₂SO₄ = n_B = 27 g/ 98 g = 0.2755 mol

Density of solution = 1.198 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.198 g cm^-3 = 83.47 cm³ = 83.47 mL = 0.08347 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2755/0.08347 = 3.301 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.2755 mol /0.073 kg = 3.774 mol L^-1

Ans: The molarity of solution is 3.374 mol L^-1 or 3.374 M, the molality of solution is 3.774 mol L^-1 or 3.774 m

Example – 06:

Calculate the mole fraction, molality and molarity of HNO₃ in a solution containing 12.2 % HNO₃. Given density of HNO₃ as 1.038 g cm^-3, H = 1, N = 14, O = 16.

Given: density of the solution = 1.038 g cm^-3, % mass of HNO₃ = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO₃ = 12.2 g and mass of H₂O = 100 – 12.2 g = 87.8 g = 0.0878 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HNO₃ = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol^-1

Number of moles of water = n_A = 87.8 g/ 18 g = 4.8778 mol

Number of moles of HNO₃ = n_B = 12.2 g/ 63 g = 0.1937 mol

Total number of moles = n_A + n_B + n_C = 4.8778 + 0.1937 = 5.0715

Mole fraction of HNO₃ = x_B = n_B/(n_A +n_B) = 0.1937/5.0715 = 0.0382

Density of solution = 1.038 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.038 g cm^-3 = 96.34 cm³ = 96.34 mL = 0.09634 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1937/0.09634 =2.011 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1937 mol /0.0878 kg = 2.206 mol kg^-1

Ans: The mole fraction of HNO3 is 0. 0382, the molarity of solution is 2.011 mol L^-1 or 2.011 M, the molality of solution is 2.206 mol kg^-1 or 2.206 m

Example – 07:

Calculate molarity and molality of 6.3 % solution of nitric acid having density 1.04 g cm^-3. Given atomic masses H = 1, N = 14 and O = 16.

Given: density of the solution = 1.04 g cm^-3, % mass of HNO₃ = 6.3 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of HNO₃ = 6.3 g and mass of H₂O = 100 – 6.3 g = 93.7 g = 0.0937 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HNO₃ = 1 g x 1 + 14 g x 1 + 16g  x 3 = 63 g mol^-1

Number of moles of water = n_A = 93.4 g/ 18 g = 5.189 mol

Number of moles of HNO₃ = n_B = 6.3 g/ 63 g = 0.1 mol

Density of solution = 1.04 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.04 g cm^-3 = 96.15 cm³ = 96.15 mL = 0.09615 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.09615 =1.040 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.1 mol /0.0937 kg = 1.067 mol kg^-1

Ans: The molarity of solution is 1.040 mol L^-1 or 1.040 M

The molality of solution is 1.067 mol kg^-1 or 1.067 m

Example – 08:

An aqueous solution of NaOH is marked 10% (w/w). The density of the solution is 1.070 g cm^-3. Calculate molarity, molality and mole fraction of NaOH in water. Given Na = 23, H =1 , O = 16

Given: density of the solution = 1.038 g cm^-3, % mass of HNO₃ = 12.2 %,

To Find: mole fraction =? molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of NaOH = 10 g and mass of H₂O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass NaOH = 23 g x 1 + 16 g x 1 + 1 g  x 1 = 40 g mol^-1

Number of moles of water = n_A = 90 g/ 18 g = 5 mol

Number of moles of NaOH = n_B = 10 g/ 40 g = 0.25 mol

Total number of moles = n_A + n_B = 5 + 0.25 = 5.25 mol

Mole fraction of NaOH = x_B = n_B/(n_A +n_B) = 0.25/5.25 = 0.0476

Density of solution = 1.070 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.070 g cm^-3 = 93.46 cm³ = 93.46 mL = 0.09346 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.25/0.09346 =2.675 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.25 mol /0.090 kg = 2.778 mol kg^-1

Ans: The molarity of solution is 2.675mol L^-1 or 2.675 M, the molality of solution is 2.778 mol kg^-1 or 2.778 m, the mole fraction of NaOH is 0. 0476

Example – 09:

A solution of glucose in water is labelled as 10 % (w/w). Calculate a) molality and b) molarity of the solution. Given the density of the solution is 1.20 g mL^-1 and molar mass of glucose is 180 g mol^-1.

Given: density of the solution = 1.20 g cm^-3, % mass of glucose = 10 %, molar mass of glucose is 180 g mol^-1.

To Find: molarity =? and molality =?

Solution:

Consider 100 g of solution

Mass of glucose = 10 g and mass of H₂O = 100 – 10 g = 90 g = 0.090 kg

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass glucose = 180 g mol^-1

Number of moles of water = n_A = 90 g/ 18 g = 5 mol

Number of moles of glucose = n_B = 10 g/ 180 g = 0.0556 mol

Density of solution = 1.20 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.20 g cm^-3 = 83.33 cm³ = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.0556/0.08333 =0.6672 M

Molality = Number of moles of solute/mass of sovent in kg

Molality = 0.0556 mol /0.090 kg = 0.6178 mol kg^-1

Ans: The molarity of solution is 0.6672 mol L^-1 or 0.6672 M, the molality of solution is 0.6178 mol kg^-1 or 0.6178 m,

Example – 10:

Battery acid 4.22 M aqueous H₂SO₄ solution, and has density 1.21 g cm^-3. What is the molality of H₂SO₄. Given H = 1, S = 32, O = 16

Given: density of the solution = 1.21 g cm^-3, Molarity of solution = 4.22 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 4.22 x 1 = 4.22

Density of solution = 1.21 g cm^-3 = 1.21 g/mL = 1.21 x 10³ g/L = 1.21 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.21 kg/L = 1.21 kg

Mass of solute (H₂SO₄) = Number of moles x molecular mass = 4.22 x 98

Mass of solute (H₂SO₄) = 413.56 g = 0.41356 kg

Mass of solvent = mass of solution – mass of solute = 1.21 – 0.41356 = 0.79644 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 4.22 mol /0.79644 kg = 5.298 mol kg^-1

Ans: Molality of solution is 5.298 mol kg^-1 or 5.298 m

Example – 11:

The density of 5.35 M H₂SO₄ solution is 1.22 g cm^-3. What is molality of a solution?

Given: density of the solution = 1.22 g cm^-3, Molarity of solution = 5.35 M.

To Find: molality =?

Solution:

Let us consider 1 L of solution

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g x 4 = 98 g mol^-1

Molarity of solution = Number of moles of the solute/volume of solution in L

Number of moles of solute = Molarity of solution x volume of solution in L = 5.35 x 1 = 5.35

Density of solution = 1.22 g cm^-3 = 1.22 g/mL = 1.22 x 10³ g/L = 1.22 kg/L

Mass of solution = Volume of solution x density = 1 L x 1.22 kg/L = 1.22 kg

Mass of solute (H₂SO₄) = Number of moles x molecular mass = 5.35 x 98

Mass of solute (H₂SO₄) = 524.3 g = 0.5243 kg

Mass of solvent = mass of solution – mass of solute = 1.22 – 0.5243 = 0.6957 kg

Molality = Number of moles of solute/mass of sovent in kg

Molality = 5.35 mol /0.6957 kg = 7.690 mol kg^-1

Ans: Molality of solution is 7.690 mol kg^-1 or 7.690 m

Example – 12:

Calculate the mole fraction of solute in its 2 molal aqueous solution.

Given: molality = 2 molal

To Find: Mole fraction =?

Solution:

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molality of solution = 2 molal = 2 mol mol kg^-1

The number of moles of solute = 2

The mass of solvent (water) = 1 kg = 1000 g

Number of moles of solvent (water) = 1000/16 = 55.55

Mole fraction of solute = 2/(2 + 55.55) = 2/57.55 = 0.03475

Ans: Mole fraction of solute is 0.0345

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Molality appeared first on The Fact Factor.

In this article, we shall study numerical problems to calculate the molarity of a given solution.

Example – 01:

A solution of NaOH (molar mass 40 g mol^-1) was prepared by dissolving 1.6 g of NaOH in 500 cm³ of water. Calculate molarity of the NaOH solution.

Given: Mass of NaOH = 1.6 g, molar mass of NaOH = 40 g mol^-1, volume of water = 500 cm³ = 500 mL = 0.5 L

To Find: Molarity =?

Solution:

Number of moles = Given mass/ Molecular mass = 1.6 g/40 g mol^-1 = 0.04 mol

Molarity = Number of moles of solute/Volume of solution in L

Molarity = 0.04 mol /0.5 L = 0.08 mol L^-1

Ans: The molarity of NaOH solution is 0.08 mol L^-1 or 0.08 M

Example – 02:

Calculate molarity of 4 g caustic soda dissolved in 200 mL of solution.

Given: Mass of solute (caustic soda) = 4 g, volume of solution = 200 mL = 0.2 L

To Find: Molarity of the solution =?

Solution:

Molar mass of caustic solute (caustic soda NaOH) = 23 g x1 + 16 g x 1 + 1 g x 1 = (23 + 16 + 1) g = 40 g

Number of moles of caustic solute (caustic soda) = given mass/molecular mass = 4 g/ 40 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

Ans: The molarity of caustic soa solution is 0.5 mol L^-1 or 0.5 M

Example – 03:

Calculate molarity of 5.3 g anhydrous sodium carbonate dissolved in 100 mL of solution.

Given: Mass of solute (sodium carbonate) = 5.3 g, volume of solution = 100 mL = 0.1 L

To Find: Molarity of the solution =?

Solution:

Molar mass of (Na₂CO₃) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of (Na₂CO₃) = given mass/molecular mass = 5.3 g/ 106 g = 0.05

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.05/0.1 = 0.5 M

Ans: The molarity of sodium carbonate solution is 0.5 mol L^-1 or 0.5 M

Example – 04:

Calculate molarity of 0.365 g pure HCl gas dissolved in 50 mL of solution.

Given: Mass of solute (HCl) = 0.365 g, volume of solution = 50 mL = 0.05 L

To Find: Molarity of the solution =?

Solution:

Molar mass of HCl = 1 g x 1 + 35.5 g x 1 = (1 + 35.5) g = 36.5 g

Number of moles of HCl = given mass/molecular mass = 0.365 g/ 36.5 g = 0.01

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.01/0.05 = 0.2 M

Ans: The molarity of HCl solution is 0.2 mol L^-1 or 0.2 M

Example – 05:

Calculate molarity of 5.85 g NaCl dissolved in 200 mL of solution.

Given: Mass of solute (NaCl) = 5.85 g, volume of solution = 200 mL = 0.2 L

To Find: Molarity of the solution =?

Solution:

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 5.85 g/ 58.5 g = 0.1

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.1/0.2 = 0.5 M

Ans: The molarity of NaCl solution is 0.5 mol L^-1 or 0.5 M

Example – 06:

Calculate molarity of 20.6 g NaBr dissolved in 500 mL of solution.

Given: Mass of solute (NaCl) = 20.6 g, volume of solution = 500 mL = 0.5 L

To Find: Molarity of the solution =?

Solution:

Molar mass of NaBr = 23 g x 1 + 80 g x 1 = (23 + 80) g = 103 g

Number of moles of NaBr = given mass/molecular mass = 20.6 g/ 103 g = 0.2

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.2/0.5 = 0.4 M

Ans: The molarity of NaBr solution is 0.4 mol L^-1 or 0.4 M

Example – 07:

Calculate molarity of pure water if its density is 1 g/mL.

Given: Density of water = 1 g/mL

To Find: Molarity of pure water =?

Solution:

Let us consider 1000 mL of water

Mass of water = volume x density = 1000 mL x 1 g/mL = 1000 g

Molar mass of water = 1 g x 2 + 16 g x 1 = (2 + 16) g = 18 g

Number of moles of water = given mass/molecular mass = 1000/ 18 g = 55.5

Molarity of pure water = Number of moles of the solute/volume of solution in L = 55.55/1 = 55.55 M

Ans: The molarity of pure water is 55.55 mol L^-1 or 55.5 M

Example – 08:

Calculate the quantity of anhydrous sodium carbonate required to produce 250 mL decimolar solution.

Given: volume of solution = 250 mL = 0.25 L, molarity = decimolar = M/10 = 0.1 M

To Find: Mass of anhydrous sodium carbonate =?

Solution:

Molarity = Number of moles of the solute/volume of solution in L

0.1 = Number of moles of the solute/0.25

Number of moles of the solute = 0.1 x 0.25 = 0.025 mol

Molar mass of (Na₂CO₃) = 23 g x 2 + 12 g x 1 + 16 g x 3 = (46 + 12 + 48) g = 106 g

Number of moles of = given mass/molecular mass

Mass of Na₂CO₃ = Number of moles x molecular mass

Mass of Na₂CO₃ = 106 x 0.025 = 2.65 g

Ans: The quantity of sodium carbonate required is 2.65 g

Example – 09:

Sulphuric acid is 95.8 % by mass. Calculate molarity and mole fraction of H₂SO₄ of density 1.91 g cm^-3. Given H = 1, S = 32, O = 16.

Given: % by mass = 95.8 %, Density of solution = 1.91 g cm^-3

To Find: Mole fraction =? Molarity =?

Solution:

Consider 100 g of solution

Mass of H₂SO₄ = 95.8 g and mass of H₂O = 100 – 95.8 g = 4.2 g

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass H₂SO₄ = 1 g x 2 + 32 g x 1 + 16g  x 4 = 98 g mol^-1

Number of moles of water = n_A = 4.2 g/ 18 g = 0.2333 mol

Number of moles of H₂SO₄ = n_B = 95.8 g/ 98 g = 0.9776 mol

Total number of moles = n_A + n_B + n_C = 0.2333 + 0.9776 = 1.2109

Mole fraction of H₂SO₄ = x_B = n_B/(n_A +n_B) = 0.9776/1.2109 = 0.8073

Density of solution = 1.91 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.91 g cm^-3 = 52.36 cm³ = 52.36 mL = 0.05236 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.9776/0.05236 = 18.67 M

Ans: The mole fraction of H₂SO₄ is 0.8073 and molarity of solution is 18.67 mol L^-1 or 18.67 M

Example – 10:

Commercially available concentrated hydrochloric acid is an aqueous solution containing 38% HCl gas by mass. If its density is 1.1 g cm^-3, calculate molarity of HCl solution and also calculate the mole fraction of HCl and H₂O.

Given: % by mass = 38 %, Density of solution = 1.1 g cm^-3

To Find: Mole fraction =? Molarity =?

Solution:

Consider 100 g of solution

Mass of HCl = 38 g and mass of H₂O = 100 – 38 g = 62 g

Molecular mass of water (H₂O) = 1 g x 2 + 16 g x 1 = 18 g mol^-1

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^-1

Number of moles of water = n_A = 62 g/ 18 g = 3.444 mol

Number of moles of HCl = n_B = 38 g/ 36.5 g = 1.041 mol

Total number of moles = n_A + n_B + n_C = 3.444 + 1.041 = 4.485

Mole fraction of HCl = x_B = n_B/(n_A +n_B) = 1.041/4.485 = 0.2321

Mole fraction of H₂O = x_A = n_A/(n_A +n_B) = 3.444/4.485 = 0.7679

Density of solution = 1.1 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.1 g cm^-3 = 90.91 cm³ = 90.91 mL = 0.09091 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.041/0.09091 = 11.45 M

Ans:  Molarity of solution is 11.45 mol L^-1 or 11.45 M, the molefraction of HCl is 0.2321 and that of H₂O is 0.7679

Example – 11:

Commercially available concentrated hydrochloric acid is an aqueous solution containing 40% HCl gas by mass. If its density is 1.2 g cm^-3, calculate molarity of HCl solution.

Given: % by mass = 40 %, Density of solution = 1.2 g cm^-3

To Find: Molarity =?

Solution:

Consider 100 g of solution

Mass of HCl = 40 g and mass of H₂O = 100 – 40 g = 60 g

Molecular mass HCl = 1 g x 1 + 35.5 g x 1 = 36.5 g mol^-1

Number of moles of HCl = n_B = 40 g/ 36.5 g = 1.096 mol

Density of solution = 1.2 g cm^-3

Volume of solution = Mass of solution / density = 100 g /1.2 g cm^-3 = 83.33 cm³ = 83.33 mL = 0.08333 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 1.096/0.08333 = 13.15 M

Ans:  Molarity of solution is 13.15 mol L^-1 or 13.15 M

Example – 12:

Calculate molarity of a solution containing 50 g of NaCl in 500 g of solution and having density 0.936 g/cm³.

Given: Mass of solute (NaCl) = 50 g, mass of solution = 500 g, density of solution = d = 0.936 g/cm³.

To Find: Molarity of solution = M =?

Molar mass of NaCl = 23 g x 1 + 35.5 g x 1 = (23 + 35.5) g = 58.5 g

Number of moles of NaCl = given mass/molecular mass = 50 g/ 58.5 g = 0.8547

Density of solution = 0.936 g cm^-3

Volume of solution = Mass of solution / density = 500 g /0.936 g cm^-3 = 534.2 cm³ = 534.2 mL = 0.5342 L

Molarity of solution = Number of moles of the solute/volume of solution in L = 0.8547/0.5342 = 1.6 M

Ans: The molarity of NaCl solution is 1.6 mol L^-1 or 1.6 M

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Concentration of Solution
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Numerical Problems on Molarity appeared first on The Fact Factor.

The concentration of a solution is the measure of the composition of a solution. For a given solution, the amount of solute dissolved in a unit volume of solution (or a unit volume of solvent) is called the concentration of the solution. It can be expressed either qualitatively or quantitatively. For example, qualitatively we can say that the solution is dilute (i.e., relatively very small quantity of solute) or it is concentrated (i.e., relatively very large quantity of solute). But in practice, it is not useful hence it is not used in chemistry. The quantitative description method gives an exact concentration of the solution and hence its concentration can be compared with the concentration of other solutions.

Methods of Expressing Concentration of the Solution Quantitatively:

Percentage by Mass or Mass Percentage (w/w):

This method is used for a solid in a liquid solution. The mass of solute in gram dissolved in the solvent to form 100 grams of the solution is called percentage by mass. The ratio of the mass of solute to the mass of the solution is called a mass fraction.

For example, if a solution is described by 10% glucose in water by mass, it means that 10 g of glucose is dissolved in 90 g of water resulting in a 100 g solution.

Concentration described by mass percentage is commonly used in industrial chemical applications. For example, a commercial bleaching solution contains a 3.62 mass percentage of sodium hypochlorite in water.

Percentage by Volume (V/V):

This method is used for liquid in a liquid solution. For example, a 10% ethanol solution in water means that 10 mL of ethanol is dissolved in 90 mL water such that the total volume of the solution is 100 mL.

Percentage by Mass by Volume (w/V):

It is the mass of solute dissolved in 100 mL of the solution. This method is commonly used in medicine and pharmacy.

Parts per million:

When a solute is present in trace quantities, it is convenient to express concentration in parts per million (ppm) and is defined as:

As in the case of percentage, concentration in parts per million can also be expressed as mass to mass, volume to volume and mass to volume.

Example: A litre of seawater (which weighs 1030 g) contains about 6 × 10^–3 g of dissolved oxygen (O₂). Such a small concentration is also expressed as 5.8 g per 10⁶ g (5.8 ppm) of seawater. The concentration of pollutants in water or atmosphere is often expressed in terms of ¼ g mL^–1 or ppm.

Strength or Concentration (Grams per litre):

It is defined as the amount of the solute in gram present in the one litre of the solution.

Mole Fraction:

The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.

It is to be noted that the sum of the mole fraction of the solute and mole fraction of liquid is 1. The concept of mole fraction is very useful in relating some physical properties of solutions, such as vapour pressure with the concentration of the solution and quite useful in describing the calculations involving gas mixtures. Mole fraction is independent of temperature

Molarity (Molar Concentration):

Molarity (M) is defined as a number of moles of solute dissolved in one litre (or one cubic decimetre) of the solution. The unit of molarity is mol L^-1 0r mol dm^-3 or M.

Number of moles of a substance can be found using the formula

Molarity changes with temperature because volume changes with temperature.

Molarity can be expressed as

• Decimolar = M/10 (0.1 M)
• Semimolar = M/2 (0.5 M)
• Pentimolar = M/5 (0.2 M)
• Centimolar = M/100 (0.01 M)
• milimolar = M/1000 (0.001 M).

Molality:

Molality (m) is defined as a number of moles of solute expressed in kg dissolved in one kg of solvent, Molality has no unit.

Molality is a better way of expressing concentration than molarity because there is no term of volume of solvent is involved. The volume of the solvent depends on the temperature of the solvent. Thus there is no effect of the change of temperature on the molality.

Molality is related to solubility as

Normality:

Normality (N) is defined as gram-equivalent of solute dissolved in one litre (or one cubic decimetre) of the solution, Unit of molarity is N.

A solution having normality equal to unity is called a normal solution.

Decinormal = N/10 (0.1 N), seminormal = N/2  (0.5 N)

Normality × equivalent mass = strength of solution in g/L.

Formality:

Formality is the number of formula mass in gram present per litre of a solution.

If the formula mass of solute is equal to its molar mass, then the formality is equal to molarity. The formality of a solution depends on temperature. This concept is used in the case of ionic substances.

A mole of an ionic compound is called formole and its molarity is called formality. Thus, the formality of a solution may be defined as a number of moles of ionic solute present in one litre of the solution.

The Relation Between Mole Fraction and Molality:

The mole fraction of any component of a solution is defined as the ratio of the number of moles of that component present in the solution to the total number of moles of all components of the solution.

Let us consider a binary solution components solvent (A) and solute (B).

Let x_A = Mole fraction of solvent

x_B = Mole fraction of solute

n_A = Number of moles of solvent

n_B = Number of moles of solute

W_A = Mass of solvent

W_B = Mass of solute

M_A = Molar mass of solvent

M_B = Molar mass of solute

Molarity of Dilution:

Molarity of Mixing:

Relation Between Molarity and Molality:

The density of a solution is in g/mL

Relation Between Molarity and Mole Fraction:

Let x_A = Mole fraction of solvent

x_B = Mole fraction of solute

n_A = Number of moles of solvent

n_B = Number of moles of solute

M = molarity of solution

d = Density of solution

M_A = Molar mass of solvent

M_B = Molar mass of solute

Mass of solution = n_AM_A  +  n_BM_B

Volume os solution = Mass of solution/density of solution

Volume os solution = (n_AM_A  +  n_BM_B)/d

The density of a solution is in g/mL

If density is in g/litre then the molarity is given as

Relation Between Normality and Molarity:

The density of a solution is in g/mL and x is the percentage of solute by mass

Related Topics

Solutions and Their Colligative Properties

• Solutions and Their Types
• Solutions of Solids and Liquids
• Numerical Problems on Percentage by Mass and Volume
• Numerical Problems on Mole Fraction
• Numerical Problems on Molarity
• Numerical Problems on Molality
• Short Cuts For Above Numerical Problems
• Solutions of Gases in Liquid
• Ideal and Non-ideal Solutions
• Lowering of Vapour Pressure
• Numerical Problems on Lowering of Vapour Pressure
• Elevation in Boiling Point and Depression in Freezing Point
• Osmosis and Osmotic Pressure

For More Topics of Chemistry Click Here

The post Concentration of Solution appeared first on The Fact Factor.