Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
2067 Upvotes
Verified Answer
The correct answer is:
$183$
$\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
$\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CaO})+\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CaCO}_3\right)\right]$
$\begin{aligned} & =[(-634)+(-393)]-[-1210] \\ & =+183 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
$\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CaO})+\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CaCO}_3\right)\right]$
$\begin{aligned} & =[(-634)+(-393)]-[-1210] \\ & =+183 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.