Q122 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
Calculate $\Delta H$ for a reaction, $2X_{(g)} + Y_{(g)} \rightarrow 2Z_{(g)}$ at $298$ K if $\Delta U = -10.5$ kJ and $R = 8.314\ \text{J K}^{-1}\ \text{mol}^{-1}$.
- A$-14.45$ kJ
- Correct answerB$-12.98$ kJ
- C$10.50$ kJ
- D$11.00$ kJ
Solution
For the given reaction $2X_{(g)} + Y_{(g)} \rightarrow 2Z_{(g)}$, the change in the number of moles of gaseous species is:
$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$
$\Delta n_g = 2 - (2 + 1) = -1$
The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is given by:
$\Delta H = \Delta U + \Delta n_g RT$
Substituting the given values:
$\Delta H = -10.5 \text{ kJ} + (-1) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298 \text{ K}$
$\Delta H = -10.5 \text{ kJ} - 2.477 \text{ kJ}$
$\Delta H = -12.977 \text{ kJ} \approx -12.98 \text{ kJ}$
Answer: $-12.98$ kJ
$\Delta n_g = \text{moles of gaseous products} - \text{moles of gaseous reactants}$
$\Delta n_g = 2 - (2 + 1) = -1$
The relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) is given by:
$\Delta H = \Delta U + \Delta n_g RT$
Substituting the given values:
$\Delta H = -10.5 \text{ kJ} + (-1) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298 \text{ K}$
$\Delta H = -10.5 \text{ kJ} - 2.477 \text{ kJ}$
$\Delta H = -12.977 \text{ kJ} \approx -12.98 \text{ kJ}$
Answer: $-12.98$ kJ
Hence, the correct answer is option: 'B'.