| Given: \[\Delta H_{f}^{o}\left( CaC{{O}_{3}},s \right)=-1207\text{ }kJ/mol;\] |
| \[\Delta H_{f}^{o}(CaO,s)=-\,635\,kJ/mol,\] |
| \[\Delta H_{f}^{o}(C{{O}_{2}},g)=-\text{ }394\text{ }kJ\text{/}mol;\] [Use\[R=8.3\text{ }J{{K}^{-1}}mo{{l}^{-1}}\]] |
A) \[702.04\text{ }kJ\]
B) \[721.96\,kJ\]
C) \[712\text{ }kJ\]
D) \[721\text{ }kJ\]
Correct Answer: A
Solution :
[a] \[CaO\left( s \right)+C{{O}_{2}}\left( g \right)-CaC{{O}_{3}}\left( s \right)\] \[{{\Delta }_{f}}{{H}^{{}^\circ }}=\Delta H_{_{f}}^{{}^\circ }\left( CaC{{O}_{3}} \right)-\Delta H_{f}^{{}^\circ }\left( CaO \right)-\Delta H_{f}^{{}^\circ }\left( CaC{{O}_{2}} \right)\]\[=-1207-\left( -635 \right)-\left( -394 \right)\] \[=-178\text{ }kJ/mol\] \[\therefore \Delta U=\Delta H-\Delta {{n}_{g}}RT\] \[\Delta U=-178\left( \frac{(-1)\times 8.3\times 300}{1000} \right)\] \[=-175.51\text{ }kJ\] \[{{n}_{CaO}}=\frac{224}{56}=4\] \[\therefore {{q}_{v}}=n{{\Delta }_{r}}U=4\times (-175.51)\] \[=-702.04\text{ }kJ\]
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