$\Delta \mathrm{S}_{\text {univ }}^{0}$ for the following reaction, at 298 K is: $\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$ $\Delta \mathrm{S}_{\mathrm{sys}}^{0}=-197 \mathrm{JK}^{-1}$, $\Delta \mathrm{H}_{\mathrm{sys}}^{0}=-91.8 \mathrm{kJ}$
(a)-197 $\mathrm{J} \mathrm{K}^{-1}$
(b)$0 \mathrm{J} \mathrm{K}^{-1}$
(c)$-308 \mathrm{J} \mathrm{K}^{-1}$
(d)$111 \mathrm{J} \mathrm{K}^{-1}$
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.