Chemical Thermodynamics Question 359

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 )-CaCO_3( s ) $

$ {\Delta _f}{{H}^{{}^\circ }}=\Delta H _{f^{\circ} }( CaCO_3 )$-$\Delta H_f^{{}^\circ }( CaO )-\Delta H_f^{{}^\circ }( CaCO_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{224}{56}=4 $

$ \therefore q_{v}=n{\Delta_{r}}U=4\times (-175.51) $

$ =-702.04kJ $



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