A) 750 K
B) 1000 K
C) 1250 K
D) 500 K
Correct Answer: A
Solution :
For the reaction, \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}X{{Y}_{3}};\Delta H=-30\,\text{kJ}\] \[\Delta {{S}^{o}}={{S}^{o}}_{(XY3)}-\left[ \frac{1}{2}S_{{{X}_{2}}}^{o}+\frac{3}{2}S_{{{Y}_{2}}}^{o} \right]\] \[=50-\left[ \frac{1}{2}\times 60+\frac{3}{2}\times 40 \right]\] \[=50-[30+60]=50-90\] \[=-40\,J{{K}^{-1}}\,mo{{l}^{-1}}\] We know that, \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\] At equilibrium, \[\Delta {{G}^{o}}=0\] \[\Delta H=T\Delta {{S}^{o}}\] \[T=\frac{\Delta H}{\Delta {{S}^{o}}}=\frac{-30\times {{10}^{3}}\text{J}}{-40\,\text{J}\,{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}}=750\,\text{K}\]
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