Q121 / 200 · MHT CET - Medical 2026 (23 April Shift 1) · Chemistry
Calculate the enthalpy of formation of HCl if bond enthalpies of $\text{H} - \text{H}$, $\text{Cl} - \text{Cl}$ and $\text{H} - \text{Cl}$ bonds are $434$ kJ mol$^{-1}$, $242$ kJ mol$^{-1}$ and $431$ kJ mol$^{-1}$, respectively.
- A$245$ kJ mol$^{-1}$
- Correct answerB$-93$ kJ mol$^{-1}$
- C$-245$ kJ mol$^{-1}$
- D$93$ kJ mol$^{-1}$
Solution
The chemical equation for the formation of $\text{HCl}$ is:
$\dfrac{1}{2}\text{H}_2(\text{g}) + \dfrac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{HCl}(\text{g})$
The enthalpy of formation is given by:
$\Delta_f H = \sum \text{Bond enthalpies of reactants} - \sum \text{Bond enthalpies of products}$
$\Delta_f H = \left[ \dfrac{1}{2} \Delta H_{\text{H-H}} + \dfrac{1}{2} \Delta H_{\text{Cl-Cl}} \right] - \Delta H_{\text{H-Cl}}$
Substituting the given values:
$\Delta_f H = \left[ \dfrac{1}{2}(434) + \dfrac{1}{2}(242) \right] - 431$
$\Delta_f H = (217 + 121) - 431$
$\Delta_f H = 338 - 431 = -93 \text{ kJ mol}^{-1}$
Answer: $-93$ kJ mol$^{-1}$
$\dfrac{1}{2}\text{H}_2(\text{g}) + \dfrac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{HCl}(\text{g})$
The enthalpy of formation is given by:
$\Delta_f H = \sum \text{Bond enthalpies of reactants} - \sum \text{Bond enthalpies of products}$
$\Delta_f H = \left[ \dfrac{1}{2} \Delta H_{\text{H-H}} + \dfrac{1}{2} \Delta H_{\text{Cl-Cl}} \right] - \Delta H_{\text{H-Cl}}$
Substituting the given values:
$\Delta_f H = \left[ \dfrac{1}{2}(434) + \dfrac{1}{2}(242) \right] - 431$
$\Delta_f H = (217 + 121) - 431$
$\Delta_f H = 338 - 431 = -93 \text{ kJ mol}^{-1}$
Answer: $-93$ kJ mol$^{-1}$
Hence, the correct answer is option: 'B'.