A) \[-\,80.8\]
B) \[-\,98.91\]
C) \[-\,50.74\]
D) \[-\,40.4\]
Correct Answer: C
Solution :
For the reaction; |
\[C+2{{H}_{2}}\xrightarrow{{}}C{{H}_{4}}\] |
Given, \[\Delta H{}^\circ =-74.81\,\,kJ\,mo{{l}^{-1}}\] |
\[\Delta S_{m}^{o}\]can be calculated as: |
\[\Delta S_{m}^{o}=S_{m(product)}^{o}-S_{m\,(reactant)}^{o}\] |
\[=S_{m}^{o}C{{H}_{4}}(g)-[S_{m}^{o}{{C}_{(graphite)}}+2S_{m}^{o}{{H}_{2}}(g)]\] |
\[=[186.3-(5.70+2\times 130.7)]J{{K}^{-1}}mo{{l}^{-1}}\] |
\[=-80.8J{{K}^{-1}}mo{{l}^{-1}}\] |
\[=-80.8\times {{10}^{-3}}J{{K}^{-1}}mo{{l}^{-1}}\] |
Since \[\Delta G{}^\circ =\Delta H{}^\circ -T\Delta S{}^\circ \] |
\[=-74.81-[(298)\times (-80.8\times {{10}^{-3}})]\] |
\[=-74.81+24.07\,\,kJ\,mo{{l}^{-1}}\] |
\[=-50.74\,kJ\,mo{{l}^{-1}}\] |
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