Q8 / 160 · TS EAMCET 2023 (13 May Shift 1) · Chemistry

Enthalpy of formation of $\mathrm{CO}(\mathrm{g}), \mathrm{CO}_2(\mathrm{~g})$ are $-110,-393$ $\mathrm{kJ} \mathrm{mol}^{-1}$ respectively. The enthalpy of combustion of CO (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is

  • Correct answer
    A
    $-283.0$
  • B
    $-110.5$
  • C
    $504$
  • D
    $-221.2$

Solution

For the combustion of $\mathrm{CO}$ :- $$ \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g}) $$ For $\mathrm{O}_2(\mathrm{~g}), \Delta \mathrm{H}_{\mathrm{f}}^{\circ}=0.0 \mathrm{~kJ} / \mathrm{mol}$. $$ \begin{aligned} & \Rightarrow \Delta \mathrm{H}_{\mathrm{comb}(\mathrm{CO})}^{\circ}=\Delta \mathrm{H}_{\mathrm{r}}^{\circ} \\ & =\Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{CO}_2\right)}^{\circ}-\left[\Delta \mathrm{H}_{\mathrm{f}(\mathrm{CO})}^{\circ}+\Delta \mathrm{H}_{\mathrm{f}\left(\mathrm{O}_2\right)}^{\circ}\right] \\ & =-393-[-110+0] \\ & =-283 \mathrm{~kJ} \mathrm{~mol}^{-1} . \end{aligned} $$

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

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