Q6 / 150 · MHT CET 2026 (18 April Shift 2) · Chemistry

Calculate the standard enthalpy change for the following reaction,
$2\text{C}_2\text{H}_6\text{(g)} + 7\text{O}_2\text{(g)} \longrightarrow 4\text{CO}_2\text{(g)} + 6\text{H}_2\text{O}\text{(l)}$
Given, $\Delta_f H^\circ(\text{C}_2\text{H}_6) = -85\ \text{kJ mol}^{-1}$
$\Delta_f H^\circ(\text{CO}_2) = -390\ \text{kJ mol}^{-1}$
$\Delta_f H^\circ(\text{H}_2\text{O}) = -285\ \text{kJ mol}^{-1}$

  • A
    $-2900\ \text{kJ}$
  • Correct answer
    B
    $-3100\ \text{kJ}$
  • C
    $-3000\ \text{kJ}$
  • D
    $-3200\ \text{kJ}$

Solution

The standard enthalpy of reaction is calculated using the formula:

$\Delta_r H^\circ = \sum \Delta_f H^\circ(\text{products}) - \sum \Delta_f H^\circ(\text{reactants})$

For the given reaction:

$\Delta_r H^\circ = [4 \times \Delta_f H^\circ(\text{CO}_2) + 6 \times \Delta_f H^\circ(\text{H}_2\text{O})] - [2 \times \Delta_f H^\circ(\text{C}_2\text{H}_6) + 7 \times \Delta_f H^\circ(\text{O}_2)]$

Substituting the given values:

$\Delta_r H^\circ = [4(-390) + 6(-285)] - [2(-85) + 7(0)]$

$\Delta_r H^\circ = [-1560 - 1710] - [-170]$

$\Delta_r H^\circ = -3270 + 170 = -3100\ \text{kJ}$

Answer: $-3100\ \text{kJ}$

Hence, the correct answer is option: 'B'.

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