Q31 / 150 · MHT CET 2022 (07 Aug Shift 1) · Chemistry

The enthalpies of combustion of C(graphite) and C(diamond) are -393.8 and $-395.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The enthalpy of conversion of $\mathrm{C}$ (graphite) to $\mathrm{C}$ (diamond) is

  • A
    $-12.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $-789.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C
    $79.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    D
    $1.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

$\begin{aligned} & \mathrm{eq}^{\mathrm{n}} 1 \ldots . . \mathrm{C}(\text { graphite })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta \mathrm{H}_1=-393.8 \frac{\mathrm{KJ}}{\mathrm{mol}} \\ & \mathrm{eq}^{\mathrm{n}} 2 \ldots . . \mathrm{C}(\text { diamond })+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) \quad \Delta \mathrm{H}_2=-395.3 \frac{\mathrm{KJ}}{\mathrm{mol}} \\ & \therefore \mathrm{eq}^{\mathrm{n}} 1-\mathrm{eq}^{\mathrm{n} 2}=\Delta \mathrm{H}_1-\Delta \mathrm{H}_2 \\ & =-393.8-(-395.3) \\ & =1.5 \frac{\mathrm{KJ}}{\mathrm{mol}}\end{aligned}$

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

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