Q21 / 150 · MHT CET 2026 (16 April Shift 2) · Chemistry
Calculate the enthalpy change when 6 g. $\text{CO(g)}$ reacts with sufficient $\text{NO}_2\text{(g)}$ according to the following reaction
$4\,\text{CO(g)} + 2\,\text{NO}_2\text{(g)} \rightarrow 4\,\text{CO}_2\text{(g)} + 2\text{N}_2\text{(g)}\,;\ \Delta_r H^0 = -1200\ \text{kJ}$
$4\,\text{CO(g)} + 2\,\text{NO}_2\text{(g)} \rightarrow 4\,\text{CO}_2\text{(g)} + 2\text{N}_2\text{(g)}\,;\ \Delta_r H^0 = -1200\ \text{kJ}$
- A$-16.08\ \text{kJ}$.
- B$-32.15\ \text{kJ}$.
- Correct answerC$-64.29\ \text{kJ}$
- D$-128.58\ \text{kJ}$.
Solution
Molar mass of $\text{CO} = 12 + 16 = 28\ \text{g mol}^{-1}$
Number of moles of $\text{CO} = \dfrac{6}{28} = \dfrac{3}{14}\ \text{mol}$
From the given balanced equation, the enthalpy change for the reaction of $4\ \text{mol}$ of $\text{CO}$ is $-1200\ \text{kJ}$.
Enthalpy change for $1\ \text{mol}$ of $\text{CO} = \dfrac{-1200}{4} = -300\ \text{kJ}$
Enthalpy change for $\dfrac{3}{14}\ \text{mol}$ of $\text{CO} = -300 \times \dfrac{3}{14} = -64.285\ \text{kJ} \approx -64.29\ \text{kJ}$
Answer: $-64.29\ \text{kJ}$
Number of moles of $\text{CO} = \dfrac{6}{28} = \dfrac{3}{14}\ \text{mol}$
From the given balanced equation, the enthalpy change for the reaction of $4\ \text{mol}$ of $\text{CO}$ is $-1200\ \text{kJ}$.
Enthalpy change for $1\ \text{mol}$ of $\text{CO} = \dfrac{-1200}{4} = -300\ \text{kJ}$
Enthalpy change for $\dfrac{3}{14}\ \text{mol}$ of $\text{CO} = -300 \times \dfrac{3}{14} = -64.285\ \text{kJ} \approx -64.29\ \text{kJ}$
Answer: $-64.29\ \text{kJ}$
Hence, the correct answer is option: 'C'.