A) \[PC{{l}_{5}}(g)\xrightarrow{{}}PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]
B) \[CaC{{O}_{3}}(s)\xrightarrow{{}}CaO(s)+C{{O}_{2}}(g)\]
C) \[N{{H}_{4}}HS(s)\xrightarrow{{}}N{{H}_{3}}(g)+{{H}_{2}}S(g)\]
D) \[{{N}_{2}}(g)+{{O}_{2}}(g)\xrightarrow{{}}2NO(g)\]
Correct Answer: C
Solution :
\[\Delta H=\Delta E+\Delta {{n}_{g}}RT\] \[\Delta H-\Delta E=\Delta {{n}_{g}}RT\] Thus, greater the value of \[\Delta {{n}_{g}},\]larger the value of\[(\Delta \Eta -\Delta E)\] [a]\[\Delta {{n}_{g}}=+1\] [b] \[\Delta {{n}_{g}}=+1\] [c] \[\Delta {{n}_{g}}=+\,2\] [d] \[\Delta {{n}_{g}}=0\]
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