In this article, we shall study to calculate the change in internal energy and change in enthalpy in a chemical reaction.

Example – 01:

For a particular reaction, the system absorbs 6 kJ of heat and does 1.5 kJ of work on its surroundings. What are the change in internal energy and enthalpy change of the system?

Given: q = + 6 kJ (Heat absorbed), W = -1.5 kJ (Work done on the surroundings).

To Find: Change in internal energy = ΔU =? Enthalpy change = ΔH =?

Solution:

By the first law of thermodynamics

Δ U = q + W

Δ U = 6 k J – 1.5 kJ = 4.5 kJ

ΔH = q_p = Heat supplied at constant pressure = + 6 kJ

Ans: The change in internal energy is 4.5 kJ and enthalpy change is 6 kJ

Example – 02:

An ideal gas expands from a volume of 6 dm³ to 16 dm³ against constant external pressure of 2.026 x 10⁵ Nm^-2. Find Enthalpy change if ΔU is 418 J.

Given:  Initial volume = V₁ = 6 dm³ = 6 × 10^-3 m³, Final volume = V₂ = 16 dm³ = 16 × 10^-3 m³, P_ext = 2.026 x 10⁵ Nm^-2, ΔU = 418 J.

To Find: Enthalpy change = ΔH =?

Solution:

Work done in the process is given by W = – P_ext × ΔV

∴ W = – P_ext × (V₂ – V₁) = – 2.026 x 10⁵ Nm^-2 × (16 × 10^-3 m³ – 6 × 10^-3 m³)

∴ W = – 2.026 x 10⁵ × (10 × 10^-3) = – 2026 J

By the first law of thermodynamics

Change in internal energy

Δ U = q_p   + W

∴   418 J = q_p – 2026 J

∴   q_p =   418 J +  2026 J = 2444 J

ΔH = q_p = Heat supplied at constant pressure = 2444 kJ

Ans: enthalpy change is 2444 J

Example – 03:

A sample of gas absorbs 4000 kJ of heat. a) If volume remains constant, what is ΔU? b) Suppose that in addition to absorption of heat by the sample, the surrounding does 2000 kJ of work on the sample. What is ΔU? c) Suppose that as the original sample absorbs heat, it expands against atmospheric pressure and does 600 kJ of work on its surroundings. What is ΔU?

Solution:

Part – a

Given: q = + 4000 kJ (Heat absorbed by sample), ΔV = 0 (Volume remains the same)

To Find: ΔU =?

Work done in the process is given by   W = – P_ext × ΔV = – P_ext × 0 = 0

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   + 0 kJ = + 4000 kJ

Part – b

Given: q = + 4000 kJ (Heat absorbed by sample), W = + 2000 kJ (Work done by surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q + W

∴ Δ U = + 4000 kJ   + 2000 kJ = + 6000 kJ

Part – c

Given: q = + 4000 kJ (Heat absorbed by sample), W = – 600 kJ (Work done on the surroundings)

To Find: ΔU =?

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = + 4000 kJ   –   600 kJ = + 3400 kJ

Example – 04:

Calculate the work done in the following reaction when 2 moles of HCl are used at constant pressure and 423 K. State whether work is on the system or by the system.

4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given: R = 8.314 J K^-1 mol^-1, T = 423 K

To Find: Work done = W =?

Solution:

The reaction is 4 HCl_(g)  + O_2(g)  →  2 Cl_2(g)   +  2 H₂O_(g)

Given 2 moles of HCl are used, hence dividing equation by 2 to get 2 HCl, we get

2 HCl_(g)  + ½O_2(g)  →   Cl_2(g)   +  H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 + 1) – (2 + ½) = 2 – 5/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 423 K = 1758 J

∴ W = + 1758 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1758 J

Example – 05:

Calculate the work done in the following reaction when 1 mol of SO₂ is oxidised at constant pressure at 5o °C. State whether work is on the system or by the system.

2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given:  Temperature = T = 5o °C = 50 + 273 = 323 K, R = 8.314 J K^-1 mol^-1,

To Find: Work done = W =?

Solution:

The reaction is 2SO_2(g) + O_2(g)  →  2 SO_3(g)

Given 1 mole of SO₂ is used, hence dividing equation by 2 to get 1 mol of SO₂

SO_2(g) + ½O_2(g)  → SO_3(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (1 ) -(1 + ½) = 1 – 3/2 = – ½

Work done in chemical reaction is given by

∴ W = – Δn RT = – (-½) mol × 8.314 J K^-1 mol^-1 × 323 K = 1343 J

∴ W = + 1343 J

Positive sign indicates that work is done by the surroundings on the system

Ans: Work done by the surroundings on the system in the reaction is 1343 J

Example – 06:

Calculate the work done in the following reaction when 2 mol of NH₄NO₃ decomposes at constant pressure at 10o °C. State whether work is on the system or by the system.

NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given:  Temperature = T = 10o °C = 100 + 273 = 373 K, R = 8.314 J K^-1 mol^-1 ,

To Find: Work done = W =?

Solution:

The reaction is NH₄NO_3(s) → N₂O_(g)  +  2 H₂O_(g)

Given 2 mol of NH₄NO₃ decomposes, hence multiplying equation by 2

2 NH₄NO_3(s) → 2 N₂O_(g)  +  4 H₂O_(g)

Δn = n_{product (g)}   – n_{reactant (g)} = (2 + 4) -(0) =6

Work done in chemical reaction is given by

∴ W = – Δn RT = – (6) mol × 8.314 J K^-1 mol^-1 × 373 K = – 18607 J

∴ W = – 18.61 kJ

Negative sign indicates that work is done by the system on the surroundings

Ans: Work done by the surroundings on the system in the reaction is – 18.61 kJ.

Example – 07:

CO reacts with O2 according to the following reaction. How much pressure volume work is done and what is the value of ΔU for the reaction of 7.0 g of CO at 1 atm pressure, if the volume change is – 2.8 L.

2CO_(g)  +  O_2(g) → 2CO_2(g)      Enthalpy change = Δ H = – 566 kJ

Given: ΔV = – 2.8 L

To Find: Work done = W =? ΔU = ?

Solution:

Work done in the process is given by

W =  – P_ext × ΔV  = – 1 atm  ×( -2.8) L = 2.8 L atm =  2.8 L atm  × 101.3 J L^-1atm^-1 = 283.6 J

∴ W = 0.2836 kJ

Positive sign indicates the work is done on the system.

From given reaction 2 x (12 + 16) = 56 g of CO on oxidation liberates  566 kJ energy

Hence heat liberated on oxidation of 7.0 g of CO = (7.0/56) × 566 = 70.75 KJ

Hence ΔH = – 70.75 kJ (negative sign as heat is liberated)

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U = 0.2836 kJ – 70.75 kJ =  -70.47 kJ

Ans: The work done on the system is 0.2836 kJ and Δ U = -70.47 kJ

Example – 08:

The enthalpy change for the following reaction is – 620 J, when 100 mL of ethylene and 100 mL of H₂ react at 1 atm pressure. Calculate pressure-volume type work and ΔU.

C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Given: ΔH = -620 J

To Find: Work done = W =? ΔU = ?

Solution:

The reaction is C₂H_4(g)  +  H_2(g) → C₂H_6(g)

Thus 1 vol of C₂H_4(g) reacts with 1 vol of H_2(g) to give 1 vol of C₂H_6(g)

Thus 100 mL of C₂H_4(g) reacts with 100 mL of H_2(g) to give 100 mL of C₂H_6(g)

Change in volume during the reaction = ΔV = V_product – V_reactant = (100) – (100 + 100) = – 100 mL = – 0.1 L

Work done in the process is given by

W = – P_ext × ΔV = – 1 atm  ×( -0.1) L = 0.1 L atm =  0.1 L atm  × 101.3 J L^-1atm^-1 = 10.13 J

∴ W = + 10.13 J

Positive sign indicates the work is done on the system.

By the first law of thermodynamics

Δ U = q   + W

∴ Δ U =   + 10.13 J – 620 J =  – 609.9 J

Ans: the work done on system is 10.13 J and Δ U = – 609.9 J

Previous Topic: The Concept of Enthalpy of a System

Next Topic: Concept of Enthalpy of a Reaction, Thermochemistry

Science > Chemistry > Chemical Thermodynamics and Energetics > Change in Internal Energy and Enthalpy

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The enthalpy of system is defined as the sum of the internal energy of the system and energy that arises due to pressure and volume. It is denoted by letter H. Mathematically,

H = U + PV

Where, H = Enthalpy U = Internal Energy P = Pressure V = Volume

As the internal energy E, Pressure P and Volume V are state functions, the enthalpy of the system is a state function.

Change in Enthalpy of System:

Let H₁ and H₂ be the enthalpies of the system in the initial state and final state respectively. Let P₁, V_1, and U1 be the pressure, volume and internal energy of the system in the initial state. Let P2, V2, and U₂ be the pressure, volume and internal energy of the system in the final state.

The change in enthalpy is given by ΔH = H₁ – H₂         ……………….    (1)      

By definition of enthalpy   H = U + PV

For initial state,       H1 = U₁ + P ₁V₁

For final state,     H₂ = U₂ + P₂ V₂

Substituting these values in equation (1)

Δ H =   (U₂ + P₂ V₂) – (U₁ + P₁V₁)   

∴ Δ H =   (U₂ – U₁) + (P₂ V₂ – P₁ V₁)

At constant pressure P₁    =   P₂ =   P 

∴ Δ H =   (U2 – U₁) + (P V₂ – P V₁) 

∴ Δ H =    (U₂ – U₁) + P (V₂ – V₁)

∴ Δ H   = ΔU + PΔ V

This is mathematical expression for change in enthalpy of system.

Thus the change in enthalpy of the system is equal to the sum of the increase in the internal energy of the system and the mechanical work due to expansion.

If Δ H is positive then the reaction is endothermic and when ΔH is negative the reaction is exothermic.

Change in Enthalpy of System at Constant Pressure (Isobaric Process):

The expression for the change in enthalpy of a system at constant pressure is

Δ H = Δ U   + PΔ V     …………  (1)

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

The mathematical statement of the first law of thermodynamics is  

ΔU = q + PΔ V 

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Let us denote q = q_p = heat absorbed by the system at constant pressure.

q_p = ΔU + PΔV    ……….   (2)  

From equations (1) and (2) we have

ΔH = q_p

Thus at constant pressure, the change in enthalpy is equal to heat absorbed at constant pressure.

Change in enthalpy of System at Constant Volume (Isochoric Process):

The expression for the change in enthalpy of a system is

Δ H = Δ U + PΔ V

Where, ΔH = Change in enthalpy ΔU = Change in internal energy P = Pressure ΔV = Change in volume

In isochoric process ΔV = 0

∴   Δ H = Δ U +  P(0)

Δ H = Δ U    ………….. (1)

The mathematical statement of the first law of thermodynamics is

Δ U = q + PΔV ………….. (2)

Where,   q = Heat supplied to the system ΔU = Change in internal energy P = Pressure ΔV = Change in volume

Substituting ΔV = 0 in equation (2)

q = ΔU

let us denote q = q_v = heat absorbed by system at constant volume.

q_v = ΔE   …………..  (3)

From equations (1) and (3) we have

ΔH = ΔU = q_v

Thus at constant volume, the change in enthalpy is equal to the heat absorbed at constant volume and also equal to change in internal energy. Thus in the isochoric process heat supplied to the system is used for increasing the internal energy of the system.

The Relation between ΔH and ΔU:

Change in the enthalpy of a system is given by

Δ H = ΔU   + PΔV ……(1)

Where, ΔU = Change in internal energy P = Pressure of the system ΔV = Change in volume

But for reactions involving gases

PV = nRT

Where P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁     = n₁RT

For Final state,    P V₂       = n₂RT

PV₂ – PV₁   =    n₂RT – n₁RT

∴ P( V₂ –  V₁)  =   (n₂ –  n₁ )RT

∴   PΔV = ΔnRT ….. (2)

Substituting in equation (1)

Δ H = ΔU   +  ΔnRT

WhereΔH =   Change in enthalpy or heat of reaction at constant pressure.

ΔU =   Change in internal energy or heat

Δn = Difference in the number of moles of gaseous products and reactants.

R   = Universal gas constant 8.314 J/ mol / k,      T   =   Absolute temperature.

Work Done in a Chemical Reaction:

The work done by a system at constant pressure and temperature is given by

W =   – P_extΔV

Assuming P_ext = P

W = – P( V₂  –  V₁)

∴ W =   – PV₂ + PV₁

But for reactions involving gases    PV = nRT

Where, P = Pressure of a gas V = Volume of the gas n = Number of moles of the gas

R = Universal gas constant T = Absolute temperature of the gas

For initial state,    P V₁ = n₁RT

For Final state,    P V₂       = n₂RT

W = – PV₂ + PV₁ =    – n₂RT + n₁RT

∴ W =   – (n₂ –  n₁ )RT

∴   W = – ΔnRT

This is an expression for work done in chemical reaction.

Conditions Under Which Change in Enthalpy of System (ΔH) is Equal to Change in Internal Energy of the System (ΔU):

When the reaction is carried out in a closed vessel, there is no change in volume. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

When the reaction involves only solids and liquids, then change in volume is negligible. ΔV = 0, hence Δ H = ΔU + PΔV gives Δ H = ΔU.

In a chemical reaction in which the number of gaseous reactants consumed is equal to the number of moles of gaseous product formed then

Δn = 0, hence ΔH =ΔU + ΔnRT gives ΔH = ΔU.

Notes: 

• Work has significance only when the number of gaseous reactants consumed and the number of moles of gaseous products formed is different and there is a change in volume.
• Heat Supplied q is not thermodynamic function but heat supplied at constant pressure (qp) and heat supplied at constant volume (qv) are thermodynamic functions:
• By the first law of thermodynamics, ΔU = q + W.  In this relation ΔU is a state function as well as thermodynamic function. Work is path function as well as non-thermodynamic function. For algebraic addition q on the right-hand side of the equation, heat supplied is path function as well as non-thermodynamic function.
• For a constant pressure process, ΔH = qp Where, ΔH is the enthalpy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant pressure qp is a thermodynamic function.
• For a constant volume process, ΔU = q_p Where, ΔU is the internal energy of a system which is state function as well as thermodynamic function. Hence heat supplied at constant volume q_v is a thermodynamic function.

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Next Topic: Numerical Problems on Enthalpy and Internal Energy Changes

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