• # question_answer Calculate the standard free energy change for the formation of methane at 298 K. The value of ${{\Delta }_{r}}H{}^\circ$ for $C{{H}_{4}}(g)$ is $-\,74.81\,kJ\,mo{{l}^{-\,1}}$and S values for C (graphite) ${{H}_{2}}(g)$ and $C{{H}_{4}}(g)$ are $5.70,$$130.7$ and $186.3J{{K}^{-\,1}}mo{{l}^{-\,1}}$ respectively. A) $-\,80.8$ B) $-\,98.91$ C) $-\,50.74$ D) $-\,40.4$

 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 $\Delta G{}^\circ =\Delta H{}^\circ -T\Delta S{}^\circ$ $=-74.81-[(298)\times (-80.8\times {{10}^{-3}})]$ $=-74.81+24.07\,\,kJ\,mo{{l}^{-1}}$ $=-50.74\,kJ\,mo{{l}^{-1}}$