Q15 / 160 · TS EAMCET 2017 · Chemistry

Given $\Delta H_f{ }^{\circ}$ for $\mathrm{CO}_2(g)$, $\mathrm{CO}(g)$ and $\mathrm{H}_2 \mathrm{O}(g)$ are $-393.5,-110.5$ and $-241.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$, respectively. The $\Delta H_f{ }^{\circ}\left[\right.$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ ] for the reaction $\mathrm{CO}_2(g)+\mathrm{H}_2(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g)$ is

  • A
    524.1
  • B
    -262.5
  • C
    -41.7
  • Correct answer
    D
    41.2

Solution

$\mathrm{CO}_2(g)+\mathrm{H}_2(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_2 \mathrm{O}(g)$ $\Delta H_f{ }^{\circ}=\langle\Delta H\rangle_{\text {products }}-(\Delta H\rangle_{\text {reactants }}$ Hence, $\Delta H_f^{\circ}=\left[\Delta H_{f_{\mathrm{CO}}}+\Delta H_{\mathrm{f}_{\mathrm{H}_2 \mathrm{O}}}\right]-\left[\Delta \mathrm{H}_{f \mathrm{CO}_2}\right]$ Given, $\Delta \mathrm{H}^{\circ}{ }_{\mathrm{f} \mathrm{CO}_2(s)}=-3935 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $$ \begin{aligned} & \Delta H_{f \mathrm{CO}}^{\circ}=-110.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta H_{f \mathrm{H}_2 \mathrm{O}}^{\circ}=-2418 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$ Put there value in equation (i) $$ \Delta H_{f^0}=-110.5-241.8+393.5=412 $$

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

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