MCQ
Bond dissociation enthalpy of $H_2, Cl_2$ and $HCl$ are $434, 242$ and $431\ kJ\, mol^{-1}$ respectively. Enthalpy of formation of $HCl$ is ............... $\mathrm{kJ \,mol}^{-1}$
- ✓$-93$
- B$245$
- C$93$
- D$-245 $
✓
Answer
Correct option: A.
$-93$
a
$\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}$
$\mathrm{H}_{2}+\mathrm{Cl}_{2} \rightarrow 2 \mathrm{HCl}$
$\Delta H_{\text {reaction }}=\Sigma(\mathrm{B.E})_{\text {reactant }}-\Sigma(\mathrm{B.E})_{\text {product }}$
$=\left[(\mathrm{B.E})_{\mathrm{H}-\mathrm{H}}+(\mathrm{B.E})_{\mathrm{Cl}-\mathrm{Cl}}\right]-\left[2 \mathrm{B.E}_{(\mathrm{H}-\mathrm{Cl})}\right]$
$=434+242-(431) \times 2$
$\Delta H_{\text {reaction }}=-186 \mathrm{kJ}$
Heat of formation is the amount of heat absorbed or evolved when one mole of substance is directly obtained from its constituent element.
Hence, enthalpy of formation of $\mathrm{HCl}=-186 / 2\; \mathrm{kJ}$
$=-93 \;\mathrm{kJ} \mathrm{mol}^{-1}$
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