Question
Standard enthalpy and standard entropy changes for the oxidation of ammonia at $$298\,K$$ are $$ - 382.64\,kJ\,mo{l^{ - 1}}$$ and $$ - 145.6\,J{K^{ - 1}}mo{l^{ - 1}},$$ respectively. Standard Gibbs energy change for the same reaction at $$298\,K$$ is
A.
$$ - 221.1\,kJ\,mo{l^{ - 1}}$$
B.
$$ - 339.3\,kJ\,mo{l^{ - 1}}$$
C.
$$ - 439.3\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 523.2\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ - 339.3\,kJ\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& \Delta {G^ \circ } = \Delta {H^ \circ } - T\Delta {S^ \circ }\,\,\,...\left( {\text{i}} \right) \cr
& {\text{Given that,}}\,\,\Delta {H^ \circ } = - 382.64\,kJ\,mo{l^{ - 1}} \cr
& \Delta {S^ \circ } = 145.6\,J\,{K^{ - 1}}\,mo{l^{ - 1}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\, = - 145.6 \times {10^{ - 3}}\,kJ\,{K^{ - 1}} \cr
& T = 298\,K \cr} $$
On putting the given values in $$eq.$$ (i) we get,
or $$\Delta {G^ \circ } = - 382.64 - $$ $$\left[ {298 \times \left( { - 145.6 \times {{10}^{ - 3}}} \right)} \right]$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = - 339.3\,kJ\,mo{l^{ - 1}}$$