Isobutane (C4H10) is burnt completely in pure oxygen as per the reaction given
below. Given that the standard heats of formation (in kcal/mole) of isobutane,
carbon dioxide, and water vapour are -31.489, -94.052, and -60.150, respectively,
the heat of reaction is ________ kcal (rounded off to 2 decimal places).
C4H10 + 6.5 O2 → 4 CO2 + 5 H2O
Official Answer Key:
-645.47
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2026 Q.52)**
**1. Intuitive Physical Setup & Given Parameters**:
Complete combustion of isobutane ($C_4H_{10}$) in oxygen:
$C_4H_{10}\text{(g)} + 6.5\,O_2\text{(g)} \to 4\,CO_2\text{(g)} + 5\,H_2O\text{(g)}$
Given standard heats of formation ($\Delta H_f^\circ$) in $\text{kcal/mole}$:
- Isobutane, $\Delta H_f^\circ(C_4H_{10}) = -31.489\text{ kcal/mole}$
- Carbon dioxide, $\Delta H_f^\circ(CO_2) = -94.052\text{ kcal/mole}$
- Water vapour, $\Delta H_f^\circ(H_2O) = -60.150\text{ kcal/mole}$
- Pure elemental oxygen in reference state, $\Delta H_f^\circ(O_2) = 0\text{ kcal/mole}$
We need to find the standard heat of reaction $\Delta H_{rxn}^\circ$ in $\text{kcal}$.
**2. Step-by-Step Thermochemical Derivation**:
1. **Hess's Law Formula**:
$\Delta H_{rxn}^\circ = \sum (\nu_i \Delta H_f^\circ)_{\text{products}} - \sum (\nu_i \Delta H_f^\circ)_{\text{reactants}}$
2. **Sum of Heats of Formation of Products**:
$\sum (\Delta H_f^\circ)_{\text{products}} = 4 \times \Delta H_f^\circ(CO_2) + 5 \times \Delta H_f^\circ(H_2O)$
$= 4 \times (-94.052) + 5 \times (-60.150)$
$= -376.208 - 300.750 = -676.958\text{ kcal}$
3. **Sum of Heats of Formation of Reactants**:
$\sum (\Delta H_f^\circ)_{\text{reactants}} = 1 \times \Delta H_f^\circ(C_4H_{10}) + 6.5 \times \Delta H_f^\circ(O_2)$
$= -31.489 + 0 = -31.489\text{ kcal}$
4. **Compute Heat of Reaction**:
$\Delta H_{rxn}^\circ = -676.958 - (-31.489) = -676.958 + 31.489 = -645.469\text{ kcal}$
Rounding to two decimal places: $-645.47\text{ kcal}$.
**3. Comprehensive Options & Distractor Analysis**:
- **Evaluation Criteria**: Numerical Answer Type (NAT).
- **Exact Analytical Result**: $-645.469\text{ kcal}$.
- **Rounding Requirement**: Two decimal places $\implies \mathbf{-645.47}$ (or magnitude $645.47$).
**4. Official Master Key & Exam Takeaway**:
- **Official Answer Key**: **-645.47** (Acceptable Range: **-646.00 to -645.00** or **645.00 to 646.00**)
- **Exam Takeaway**: By Hess's law, heat of combustion is $\sum \nu_p \Delta H_{f,p} - \sum \nu_r \Delta H_{f,r}$. Exothermic reactions have $\Delta H < 0$.
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