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NVAMediumJEE 2025Hess's Law

JEE Chemistry 2025 Question with Solution

Consider the following cases of standard enthalpy of reaction (ΔHf∘\Delta H_f^\circ in kJ mol−1\text{kJ mol}^{-1}):

C2H6(g)+7 O2(g)→2 CO2(g)+3 H2O(l)ΔH1∘=−1550\text{C}_2\text{H}_6(g) + 7 \, \text{O}_2(g) \rightarrow 2 \, \text{CO}_2(g) + 3 \, \text{H}_2\text{O}(l) \quad \Delta H_1^\circ = -1550 C(graphite)+O2(g)→CO2(g)ΔH2∘=−393.5\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2^\circ = -393.5 H2(g)+12 O2(g)→H2O(g)ΔH3∘=−286\text{H}_2(g) + \frac{1}{2} \, \text{O}_2(g) \rightarrow \text{H}_2\text{O}(g) \quad \Delta H_3^\circ = -286

The magnitude of ΔHf∘\Delta H_f^\circ of C2H6(g)\text{C}_2\text{H}_6(g) is _____ kJ mol−1\text{kJ mol}^{-1} (Nearest integer).

Answer

Correct answer:94

Step-by-step solution

Standard Method

Given:

  • Combustion of C2H6(g)\text{C}_2\text{H}_6(g):
C2H6(g)+7 O2(g)→2 CO2(g)+3 H2O(l)\text{C}_2\text{H}_6(g) + 7 \, \text{O}_2(g) \rightarrow 2 \, \text{CO}_2(g) + 3 \, \text{H}_2\text{O}(l)

with ΔH1∘=−1550 kJ mol−1\Delta H_1^\circ = -1550 \, \text{kJ mol}^{-1}.

  • Formation of CO2(g)\text{CO}_2(g):
C(graphite)+O2(g)→CO2(g)\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g)

with ΔH2∘=−393.5 kJ mol−1\Delta H_2^\circ = -393.5 \, \text{kJ mol}^{-1}.

  • Formation of H2O(l)\text{H}_2\text{O}(l) used in the working:
H2(g)+12 O2(g)→H2O(l)\text{H}_2(g) + \frac{1}{2} \, \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l)

with ΔHf∘=−285.8 kJ mol−1\Delta H_f^\circ = -285.8 \, \text{kJ mol}^{-1}.

Find: The magnitude of ΔHf∘\Delta H_f^\circ of C2H6(g)\text{C}_2\text{H}_6(g).

Using Hess's law for the combustion reaction,

ΔHcomb=2ΔHf∘(CO2)+3ΔHf∘(H2O(l))−ΔHf∘(C2H6)\Delta H_{\text{comb}} = 2\Delta H_f^\circ(\text{CO}_2) + 3\Delta H_f^\circ(\text{H}_2\text{O}(l)) - \Delta H_f^\circ(\text{C}_2\text{H}_6)

So,

ΔHf∘(C2H6)=2ΔHf∘(CO2)+3ΔHf∘(H2O(l))−ΔHcomb\Delta H_f^\circ(\text{C}_2\text{H}_6) = 2\Delta H_f^\circ(\text{CO}_2) + 3\Delta H_f^\circ(\text{H}_2\text{O}(l)) - \Delta H_{\text{comb}}

Substituting the values,

ΔHf∘(C2H6)=2(−393.5)+3(−285.8)−(−1550)\Delta H_f^\circ(\text{C}_2\text{H}_6) = 2(-393.5) + 3(-285.8) - (-1550)

Now evaluate stepwise,

2(−393.5)=−787.02(-393.5) = -787.0 3(−285.8)=−857.43(-285.8) = -857.4 −787.0+(−857.4)=−1644.4-787.0 + (-857.4) = -1644.4 −1644.4−(−1550)=−1644.4+1550=−94.4 kJ mol−1-1644.4 - (-1550) = -1644.4 + 1550 = -94.4 \, \text{kJ mol}^{-1}

Therefore,

ΔHf∘(C2H6(g))=−94.4 kJ mol−1\Delta H_f^\circ\big(\text{C}_2\text{H}_6(g)\big) = -94.4 \, \text{kJ mol}^{-1}

Hence, the magnitude is 94.4 kJ mol−194.4 \, \text{kJ mol}^{-1}, and the nearest integer is 9494.

The solution also notes that an earlier value of 84.484.4 is an arithmetic error; the correct value from the shown numbers is 94.494.4.

Target Formation Reaction

Given: The target formation reaction is

2C(graphite)+3H2(g)→C2H6(g)2\text{C(graphite)} + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)

Find: Its standard enthalpy of formation.

From the provided working, first form the products of combustion from the elements:

2C(graphite)+2O2(g)→2CO2(g)2\text{C(graphite)} + 2\text{O}_2(g) \rightarrow 2\text{CO}_2(g)

with enthalpy

2×(−393.5)=−787.0 kJ mol−12 \times (-393.5) = -787.0 \, \text{kJ mol}^{-1}

Also,

3H2(g)+32O2(g)→3H2O(l)3\text{H}_2(g) + \frac{3}{2}\text{O}_2(g) \rightarrow 3\text{H}_2\text{O}(l)

with enthalpy

3×(−285.8)=−857.4 kJ mol−13 \times (-285.8) = -857.4 \, \text{kJ mol}^{-1}

Adding these gives

2C(graphite)+3H2(g)+72O2(g)→2CO2(g)+3H2O(l)2\text{C(graphite)} + 3\text{H}_2(g) + \frac{7}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l)

with

ΔH=−1644.4 kJ mol−1\Delta H = -1644.4 \, \text{kJ mol}^{-1}

Now reverse the combustion of ethane:

2CO2(g)+3H2O(l)→C2H6(g)+72O2(g)2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \rightarrow \text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g)

with

ΔH=+1550 kJ mol−1\Delta H = +1550 \, \text{kJ mol}^{-1}

Adding both equations,

2C(graphite)+3H2(g)→C2H6(g)2\text{C(graphite)} + 3\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g)

Thus,

ΔHf∘=−1644.4+1550=−94.4 kJ mol−1\Delta H_f^\circ = -1644.4 + 1550 = -94.4 \, \text{kJ mol}^{-1}

So the required magnitude is 9494 to the nearest integer.

Common mistakes

  • Using H2O(g)\text{H}_2\text{O}(g) directly from the question instead of matching the combustion product H2O(l)\text{H}_2\text{O}(l) is incorrect, because Hess's law requires the same physical state in the enthalpy balance. Use the liquid-water value shown in the solution working.

  • Forgetting that the question asks for the magnitude is incorrect, because the calculated enthalpy of formation is negative but the final numerical answer must be positive. After finding −94.4-94.4, report 9494 as the nearest integer.

  • Making the arithmetic error 1550−787−857.4=−84.41550 - 787 - 857.4 = -84.4 is incorrect, because the correct subtraction gives −94.4-94.4. Add carefully or combine −787-787 and −857.4-857.4 first before adding 15501550.

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