Question
In which of the following reactions, standard reaction entropy changes $$\left( {\Delta {S^ \circ }} \right)$$ is positive and standard Gibbs energy change $$\left( {\Delta {G^ \circ }} \right)$$ decreases sharply with increasing temperature?
A.
$$C\left( {{\text{graphite}}} \right) + \frac{1}{2}{O_2}\left( g \right) \to CO\left( g \right)$$
B.
$$CO\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to C{O_2}\left( g \right)$$
C.
$$Mg\left( s \right) + \frac{1}{2}{O_2}\left( g \right) \to MgO\left( s \right)$$
D.
$$\frac{1}{2}C\left( {{\text{graphite}}} \right) + \frac{1}{2}{O_2}\left( g \right) \to \frac{1}{2}C{O_2}\left( g \right)$$
Answer :
$$C\left( {{\text{graphite}}} \right) + \frac{1}{2}{O_2}\left( g \right) \to CO\left( g \right)$$
Solution :
Among the given reactions only in the case of
$$C\left( {{\text{graphite}}} \right) + \frac{1}{2}{O_2}\left( g \right) \to CO\left( g \right)$$
entropy increases because randomness (disorder)
increases. Thus, standard entropy change $$\left( {\Delta {S^ \circ }} \right)$$ is
positive. Moreover, it is a combustion reaction and
all the combustion reactions are generally
exothermic, i.e. $$\Delta {H^ \circ } = - ve$$
We know that,
$$\eqalign{
& \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ } \cr
& \Delta {G^ \circ } = - ve - T\left( { + ve} \right) \cr} $$
Thus, as the temperature increases, the value of $$\Delta {G^ \circ }$$ decreases.