Q11 / 160 · AP EAMCET 2023 (18 May Shift 1) · Chemistry

The standard enthalpy of formation of $\mathrm{CO}_2(\mathrm{~g})$, $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CaCO}_3(\mathrm{~s})$ are $-393,-634,-1210 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If all the substances are in standard state, the standard enthalpy of decomposition of calcium carbonate to $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CO}_2$ (g) (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is

  • A
    $969$
  • Correct answer
    B
    $183$
  • C
    $-969$
  • D
    $-183$

Solution

$\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$ $\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CaO})+\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CaCO}_3\right)\right]$ $\begin{aligned} & =[(-634)+(-393)]-[-1210] \\ & =+183 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

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

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