# Problems on Enthalpy Change and Internal Energy Change

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### Sign Conventions for q, W, and ΔU:

• When heat is absorbed by the system q is positive.
• When heat is rejected or given out by the system q is negative.
• When the work is done on the system by surroundings (Work of compression) then W is positive.
• When the work is done by the system on the surroundings (Work of expansion) then, W is negative.
• When there is an increase in internal energy of the system ( increase in temperature) ΔU is positive.
• When there is a decrease in internal energy of the system ( decrease in temperature) ΔU is negative.

### Problems based on Enthalpy Change and Internal Energy Change in Chemical reaction:

#### Example – 1:

• 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?
• Solution:
• 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 = ?

By the first law of thermodynamics

Δ U  =  q   + W

Δ U  =  6 k J – 1.5 kJ = 4.5 kJ

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

Hence change in internal energy is 4.5 kJ and enthalpy  change is 6 kJ

#### Example – 2:

• 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.
• Solution:
• Given:  Initial volume = V1 = 6 dm³ = 6 × 10-3 m³, Final volume =  V2 = 16 dm³ = 16 × 10-3 m³, Pext = 2.026 x 105 Nm-2, ΔU = 418 J.
• To Find: Enthalpy change = ΔH = ?

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

∴  W = – Pext × (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

Δ U  =  qp   + W

∴   418 J  =  qp – 2026 J

∴   qp =   418 J +  2026 J = 2444 J

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

Hence enthalpy change is 2444  J

#### Example – 3:

• 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  =  – Pext × ΔV  =  – Pext × 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 – 4:

• 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)  + O2(g)  →  2 Cl2(g)   +  2 H2O(g)

• Solution:.
• Given: R =  8.314 J K-1 mol-1 , T = 423 K
• To Find: Work done = W = ?

The reaction is 4 HCl(g)  + O2(g)  →  2 Cl2(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)  + ½O2(g)  →   Cl2(g)   +  H2O(g)

Δn =  nproduct (g)   – nreactant (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

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

#### Example – 5:

• 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.

2SO2(g) + O2(g)  →  2 SO3(g)

• Solution:
• Given:  Temperature = T = 5o °C = 50 + 273 = 323 K , R =  8.314 J K-1 mol-1 ,
• To Find: Work done = W = ?

The reaction is 2SO2(g) + O2(g)  →  2 SO3(g)

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

SO2(g) + ½O2(g)  → SO3(g)

Δn =  nproduct (g)   – nreactant (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

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

#### Example – 6:

• Calculate the work done in the following reaction when 2 mol of NH4NO decomposes at constant pressure at 10o °C. State whether work is on the system or by the system.

NH4NO3(s) → N2O(g)  +  2 H2O(g)

• Solution:
• Given:  Temperature = T = 10o °C = 100 + 273 = 373 K , R =  8.314 J K-1 mol-1 ,
• To Find: Work done = W = ?

The reaction is NH4NO3(s) → N2O(g)  +  2 H2O(g)

Given 2 mol of NH4NO decomposes , hence multiplying equation by 2

2 NH4NO3(s) → 2 N2O(g)  +  4 H2O(g)

Δn =  nproduct (g)   – nreactant (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

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

#### Example – 7:

• 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)  +  O2(g) → 2CO2(g)      Enthalpy change = Δ

H =  – 566 kJ

• Solution:
• Given: ΔV = – 2.8 L
• To Find: Work done = W = ?, ΔU = ?

Work done in the process is given by

W  =  – Pext × 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

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

Hence work dobe on system is 0.2836 kJ and  Δ U =  -70.47 kJ

#### Example – 8:

• The enthalpy change for 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.

C2H4(g)  +  H2(g) → C2H6(g)

• Solution:
• Given: ΔH = -620 J
• To Find: Work done = W = ?, ΔU = ?

The reaction is C2H4(g)  +  H2(g) → C2H6(g)

Thus 1 vol of C2H4(g) reacts with 1 vol of H2(g) to give 1 vol of C2H6(g)

Thus 100 mL of C2H4(g) reacts with 100 mL of H2(g) to give 100 mL of C2H6(g)

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

Work done in the process is given by

W  =  – Pext × 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

By the first law of thermodynamics

Δ U  =  q   + W

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

Hence work dobe on system is 10.13 J and  Δ U = – 609.9 J

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