A) \[\text{C}\,\text{(graphite)}+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g)\]
B) \[CO(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}C{{O}_{2}}(g)\]
C) \[Mg(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}MgO(s)\]
D) \[\frac{1}{2}C(\text{graphite})+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}\frac{1}{2}C{{O}_{2}}(g)\]
Correct Answer: A
Solution :
Among the given reactions only in the case of C (graphite)\[+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}CO(g)\]entropy increases because randomness (disorder) increases. Thus, standard entropy change \[(\Delta {{S}^{o}})\] is positive. Moreover, it is a combustion reaction and all the combustion reactions are generally exothermic, i.e.,\[\Delta {{H}^{o}}=-ve\] We know that \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\] \[\Delta {{G}^{o}}=-ve-T(+ve)\] Thus, as the temperature increases, the value of \[\Delta {{G}^{o}}\] decreases.
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