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  • In a container with fixed volume, calcium carbonate was formed by the reaction between

In a container with fixed volume, calcium carbonate was formed by the reaction between 242 \mathrm{~g} of calcium oxide with carbon dioxide at 27^{\circ} \mathrm{C}. Calculate theheat produced in the same if 
\mathrm{\begin{aligned} & \Delta H_{f\left(C_a \mathrm{CO}_3, 5\right)}=-1205 \mathrm{~kJ} / \mathrm{mol} \\ & \Delta H_{f \dot{(C_{a}O,5)}}=-640 \mathrm{~kJ} / \mathrm{mol} \\ & \end{aligned}}
\mathrm{\Delta H_{f\left(\mathrm{CO}_2, g\right)}^{\circ}=-400 \mathrm{~kJ} / \mathrm{mol}}

Option: 1

\mathrm{702.2\ kJ}


Option: 2

\mathrm{721.96\ kJ}


Option: 3

\mathrm{712\ kJ}


Option: 4

\mathrm{721\ kJ}


Answers (1)

best_answer

\begin{gathered} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(g) \rightarrow \mathrm{CaCO}_3(\mathrm{~s}) \\ \Delta H_f^{\circ}=\Delta H_f^{\circ}\left(\mathrm{CaCO}_3\right)-\Delta H_f^{\circ}(\mathrm{CaO})-\Delta H_f^{\circ}\left(\mathrm{CO}_2\right) \end{gathered}
\mathrm{\begin{aligned} & =-1205+640+400 \\ & =-165 \mathrm{~kJ} / \mathrm{mol} \end{aligned}}
\mathrm{\begin{aligned} \Delta V & =\Delta H-\Delta_{n g} R T \\ & =-165-\frac{(-1) \times 8.3 \times 300}{1000} \end{aligned}}
\mathrm{\begin{aligned} & =-162.51 \\ n_{\text {CaO }} & =\frac{242}{56}=4.32 \end{aligned}}
\mathrm{\begin{aligned} \therefore \mathrm{z_v} & =n \Delta u \\ & =4.32(-162.51) \\ & =-702.2 \mathrm{~kJ} \end{aligned}}

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