A) \[Mg(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}MgO(s)\]
B) \[\frac{1}{2}C(graphite)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}\frac{1}{2}C{{O}_{2}}(g)\]
C) \[C(graphite)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g)\]
D) \[CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}C{{O}_{2}}(g)\]
Correct Answer: C
Solution :
\[C(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CO(g);\Delta n+\frac{1}{2}.\] Also the moles of gases increase and therefore entropy change \[(\Delta S)\] is positive. An increase in temperature will cause more change in \[T\Delta S.\]Also it is a combustion reaction and thus \[\Delta \Eta =-ve\] Since \[\Delta G=\Delta H=T\Delta S\] \[=-ve-(+ve)=-ve\]
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