Characteristics of Equilibrium Constant:

• It has a definite value for every chemical reaction at a particular temperature.
• It is independent of the initial concentrations of the reacting species.
• It changes with the change in the temperature.
• It depends on the nature of the reaction.
• It is independent of the change of pressure, volume and concentrations of the reactants and products.
• It is not affected by the introduction of the catalyst.
• The expression for it may contain the concentrations of gases or molecules and ions in solution but not of pure solids or pure liquids.
• The expression for it and its magnitude depends on the stoichiometric form of the balanced chemical equation.
• When the equation for equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.

If K_c is equilibrium constant for reaction aA + bB ⇌  cC + dD

Then its value for reaction naA + nbB ⇌  ncC + ndD is given by

K’_c = (K_c)ⁿ

• When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the value of K_C of the resultant equilibrium.

K_(resultant) = K_{(Reaction 1)}  x K_{(Reaction 2)}

• If K₁, k₂, K₃, …. are equilibrium constant for recation₁, reaction₂, reaction₃, ……. Then the value of K_C for reaction, a x recation₁+ b x reaction₂, c x reaction₃, ……. is given by

K_C = (K₁)^a(K₂)^b(K₃)^c……..

• For a reversible reaction, the value of K_C for the backward reaction is inverse of the equilibrium constant for the forward reaction

K_(backward) = 1 / K_(forward)

• If it is expressed in terms of concentration, it has different units for different reactions.

K_C is Dimensionless Unitless Quantity:

Depending upon the stoichiometric coefficients of a chemical reaction, the equilibrium constant should have a unit. Its unit should be

(mol dm^-3)^{(∑n products – ∑n reactants)}

Note that 1 dm³ = 1 L. Hence concentration can be expressed as (mol L^-1)

These days we specify equilibrium constant in terms of dimensionless quantities by specifying the standard state of reactants and products. The standard state pressure for pure gas is 1 bar and the partial pressures of the gases are measured with respect to this standard state.

If a gas has a partial pressure of 1.5 bar, then in terms of standard state its pressure would be equal to 1.5 bar/1bar = 1.5, a dimensionless number. Similarly, for concentrations, the standard state is 1 M ( 1mol dm^-3). If the concentration of the substance is 2.0 M. Then in terms of standard state it will be expressed as 2.0 M/1 M = 2. Thus using partial pressures and concentrations K_P and K_C obtained are dimensionless. Hence equilibrium constant is considered as dimensionless, unitless quantity.

K_C of a General Reaction and its Multiple:

Consider a hypothetical reversible reaction

aA + bB ⇌  cC + dD

The equilibrium constant of the reaction given by

Consider a multiple of above reaction

naA + nbB ⇌  ncC + ndD

The equilibrium constant of the reaction given by

Thus when the equation for an equilibrium is multiplied by a factor, then the equilibrium constant must be raised to the power equal to the factor.

K_C for the addition of Two Chemical Equilibria:

Consider following equilibria

1) N_2(g) +   O_2(g)  ⇌     2NO_(g)

The equilibrium constant for the reaction is

2)   2NO_(g) +   O_2(g)  ⇌     2NO_2(g) 

The equilibrium constant for the reaction is

3)  N_2(g) +  2O_2(g)   ⇌   2NO_2(g)

The equilibrium constant for the reaction is

Multiplying equation (1) by (2) we get

Thus When the addition of two equilibria leads to another equilibrium then the product of their equilibrium constants gives the equilibrium constant of the resultant equilibrium.

Temperature Dependence of Equilibrium Constant:

At chemical equilibrium, the rate of the forward reaction is equal to the rate of backward reaction. When the temperature is increased, in general, the rate of both the forward reaction and the backward reaction increases. As the energy of activation of the forward reaction and the backward reaction are different, the extent of the increase of the forward reaction and the backward reaction is different. Thus the value of K_f (rate constant for the forward reaction) and K_b (rate constant for the backward reaction) changes to a different extent. Thus the ratio K_f / K_b changes. i.e. the value of equilibrium constant changes with the change in temperature.

The value of the equilibrium constant for endothermic reaction increases with an increase in temperature, while the value of the equilibrium constant for exothermic reaction decreases with an increase in temperature. The temperature dependence of equilibrium constant can be written mathematically as

Uses of K_C:

It helps in the prediction of the extent of reaction:

The magnitude of the K_C tells us about the extent in which the reactants are converted into the products before the equilibrium is attained. Larger values of K indicates that the extent of reactants converting into products is greater. The generalization is

• If K_C > 10³ Products predominates the reactants. i.e. the concentration of products is very high compared to that of reactants t equilibrium and the reaction proceeds nearly to completion
• If K_C < 10^-3 Reactants predominates the products. i.e. the concentration of products is very less compared to that of reactants t equilibrium and the reaction hardly proceeds.
• 10^-3 < K_C < 10^-3 , appreciable concentrations of both the reactants and products are present at equilibrium

It gives us an idea of relative stabilities of reactants and products:

If the K_C is large products are more stable than the reactants. Whereas, If the value of the equilibrium constant is small reactants are more stable than the products.  The generalisation is

If K_C > 10³ Products are stable than reactants.

If K_C < 10^-3 Reactants are stable than products.

It helps in the prediction of the direction of a net reaction.

The value of K_C helps in the prediction of the direction in which the net reaction is proceeding at given concentrations or partial pressures of reactants and products. If Q_c is the concentration quotient and K_c be the equilibrium constant of a chemical reaction.

For reaction aA + bB ⇌  cC + dD

When computing Q_C the concentration at that instant are used

When computing K_C the concentration at equilibrium are used. The generalization is

• If Q_C > K_C , the reaction is taking place in a backward direction i.e. in the direction of reactants
• If Q_C <  K_C , the reaction is taking place in a forward direction i.e. in the direction of products.
• If Q_C =  K_C The reaction is in the equilibrium state and hence no net reaction is taking place.

It helps in the calculation of equilibrium constants and equilibrium pressures. 

If the equilibrium concentrations of various reactants and products are known for a reaction, the equilibrium constant can be calculated. On the other hand, if the equilibrium constant is known, the equilibrium concentrations can be calculated.

Steps Involved in the Calculation of K_C and Equilibrium Pressures:

1. Write the chemical equation for the equilibrium
2. Write expression for K_C or K_P for the reaction.
3. Express all unknown concentrations or partial pressures in terms of a single variable x.
4. Substitute equilibrium concentrations or partial pressures in terms of x in the expression for K_C or K_P.
5. Solve the equation for x
6. Substitute the value obtained for x in the expression in step 3 to calculate equilibrium concentrations or equilibrium partial pressures

Science > Chemistry > Chemical Equilibrium > Equilibrium Constant

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In this article, we shall stuudy to write expression for equilibrium constant.

Steps Involved in Writing Expression for Equilibrium Constant of a Reaction:

• Write the balanced chemical equation for the reaction.
• Write the products of equilibrium concentrations of the products in the numerator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Write the products of equilibrium concentrations of the reactants in the denominator. Omit pure solids, pure liquids and the solvents in dilute solutions.
• Raise each concentration term to the power equal to stoichiometric coefficients of the species in the equation.

Homogeneous Equilibrium:

The equilibrium in which all the substances involved exist in a single homogeneous phase is called homogeneous equilibrium.

Examples:

N_2(g) +  3 H_2(g)  ⇌     2NH_3(g)

H_2(g) +  I_2(g) ⇌     2HI_(g)

2SO_2(g) +  O_2(g)  ⇌   2SO_3(g)

NH_3(aq) + H₂O_(l) ⇌     2NH₄+_(aq) +  OH-_(aq)

2N₂O_(g) ⇌ 2N_2(g)  +   O_2(g)

CH₃COOC₂H_5(aq) + H2O_(l) ⇌ CH₃COOH_(aq) +   C2H5OH_(aq)

Heterogeneous Equilibrium:

The equilibrium in which the substance involved are present in different phases is called heterogeneous equilibrium.

Examples:

CaCO_3(s) ⇌  CaO_(s) + . CO_2(g)

CaCO_3(s)+H₂O(l)+CO₂(g) ⇌ Ca²⁺_(aq)+ 2HCO₃^–_(aq)

(NH₄)₂CO_3(s) ⇌ 2NH_3(g) + CO_2(g)+  H₂O_(g)

2Mg_(s) + O2_(g)  ⇌    2MgO_(s)

Note: Pure liquids and solids are ignored while writing the equilibrium constant expression.

Now, for pure solids and pure liquids, the molar mass M and density ρ are constant. Hence the concentration remains constant.

• When pure solids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all solids involved in equilibrium are taken as unity.
• When pure liquids are involved in equilibrium, its concentration remains constant no matter how much of it is present. Therefore by convention, the concentrations of all liquids involved in equilibrium are taken as unity.
• For equilibria in the aqueous medium, the concentration of solvent (water) will not change appreciably because it is present in large excess. Hence by convention, the concentration of solvent (water) is taken as unity.
• Hence in general, pure liquids, pure solids, and solvents can be ignored while writing the equilibrium constant expression.

Writing the Expression for K_c and K_p:

BaCO_3(s)  ⇌  BaO_(s) + CO_2(g)

4NH_3(g) + 5O_2(g) ⇌   4NO_(g)+ 6H₂O_(g)

NH_3(g) + HCl_(g)  ⇌  NH₄Cl_(s)

3Cl_2(g) + 2NO_2(g)  ⇌  2NO₂Cl_3(g)

Fe³⁺_(aq) + SCN^–_(aq) ⇌ FeNCS²⁺_(aq)

CH_4(g) + 2H₂S_(g)  ⇌  CS_2(g)+ 4H_2(g)

MgCO_3(s)  ⇌  MgO_(s)  + CO_2(g)

AgBr_(s) ⇌ Ag⁺_(aq)+ Br^–_(aq)

K_c = [Ag⁺][Br ^–]

CH₃COCH_3(l) ⇌CH₃COCH_3(g)

K_c = [CH₃COCH_3(g)]

CH_4(g) + 2O_2(g) ⇌ CO_2(g)+ 2H₂O_(g)

Al_(s) + 3H⁺_(aq) ⇌ Al³⁺_(aq)+ 3/2H_2(g)

HPO₄^2-_(aq) + H₂O_(l) ⇌ H₃O⁺_(aq)+ PO₄^3-_(aq)

Ag₂O_(s) + 2HNO_3(aq) ⇌  2AgNO_3(aq)+ H₂O_(l)

Ni_(s) + 4CO_(g) ⇌  Ni(CO)_4(g)

CuO_(s) + H_2(g) ⇌  Cu_(s)+ H₂O_(g)

N_2(g) + 3H_2(g) ⇌  2NH_3(g)

Fe³⁺_(aq) + 3OH^–_(aq) ⇌  Fe(OH)_3(s)

2N₂O_(g) ⇌ 2N_2(g) + O_2(g)

C_(s)+  CO_2(g) ⇌  2CO_(g)

Consider a hypothetical reversible reaction involving homogeneous gaseous phase

aA(g) + bB(g) ⇌  cC(g) + dD(g)

The equilibrium constant in terms of the partial pressure is given by

This is the relation between K_c and K_p.

Where Δn = (Sum of the exponents in the numerator of concentration quotient) – (Sum of the exponents in the denominator of concentration quotient)

Steps Involved in Finding Relation Between K_c and K_p :

• Write the balanced chemical equation for the reaction.
• Find the change in the number of moles of gaseous species by the following formula.

• Now, use the relation

Examples to Find Relation between Between K_c and K_p :

Example – 01:

Example – 02:

Example – 03:

Example – 04: 

For the reaction N_2(g) + 3H_2(g) ⇌  2NH_3(g), the value of the equilibrium constant K_p is 3.6 x 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 05:

For the reaction 2SO_2(g)) + O_2(g) ⇌ 2SO_3(g), the value of the equilibrium constant K_p is 2.0 X 10¹⁰ bar^-1 at 450 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 06:

For the reaction 2NOCl_(g) ⇌ 2NO_(g)+ Cl_2(g), the value of the equilibrium constant K_p is1.8 X 10^-2 at 500 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 07:

For the reaction CaCO_3(s)  ⇌ CaO_(s)+ CO_2(g), the value of equilibrium constant K_p is 167 at 1073 K. Calculate the value of K_c for the reaction at the same temperature. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 08:

Find the ratio of K_p/K_c for the reaction, CO(g) + 1/2O_2(g)  ⇌  2CO_2(g), at 500 K. Take R = 0.0831 bar lit mol^-1 K^-1.

Example – 09:

Find the ratio of K_p/K_c  for the reaction, N_2(g) + O_2(g) ⇌   2NO_(g)

Example – 10:

The equilibrium constant K_p for the reaction H_2(g)  + I_2(g)  ⇌  2HI_(g) is 130 at 510 K. Calculate K_c for following reactions a) 2HI_(g)  ⇌ H_2(g)  + I_2(g),  b) HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g), c) 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

∴  K_c = 130

For the reaction 2HI_(g)  ⇌ H_2(g)  + I_2(g),

For the reaction, HI_(g)  ⇌ 1/2H_2(g)  + 1/2I_2(g)

For reaction, 1/2H_2(g)  + 1/2I_2(g)  ⇌  HI_(g).

Science > Chemistry > Chemical Equilibrium > Writing Expression for Equilibrium Constant

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