Chemical Thermodynamics Question 211
Question: Calculate the heat produced (in kJ) when 224 g of $ CaO $ is completely converted to $ CaCO_3 $ by reaction with $ CO_2 $ at $ 27{}^\circ C $ in a container of fixed volume.
Given: $ \Delta H _{f}^{o}( CaCO_3,s )=-1207kJ/mol; $
$ \Delta H _{f}^{o}(CaO,s)=-,635,kJ/mol, $
$ \Delta H _{f}^{o}(CO_2,g)=-394kJ\text{/}mol; $ [Use $ R=8.3J{{K}^{-1}}mo{{l}^{-1}} $ ]
Options:
A) $- 702.04kJ $
B) $ 721.96,kJ $
C) $ 712kJ $
D) $ 721kJ $
Show Answer
Answer:
Correct Answer: A
Solution:
$ CaO( s )+CO_2( g )\rightarrow CaCO_3( s ) $
$ {\Delta _f}{{H}^{{}^\circ }}=\Delta H _{f^{\circ} }( CaCO_3 )$-$\Delta H_f^{{}^\circ }( CaO )-\Delta H_f^{{}^\circ }( CO_2 ) $
$ =-1207-( -635 )-( -394 ) $
$ =-178kJ/mol $
$ \therefore \Delta U=\Delta H-\Delta n_{g}RT $
$ \Delta U=-178( \frac{(-1)\times 8.3\times 300}{1000} ) $
$ =-175.51kJ $
$ n_{CaO}=\frac{22.4}{56}=0.4 $
$ \therefore q_{v}=n{\Delta_{r}}U=4\times (-175.51) $
$ = -702.04\ \text{kJ}$
BETA
