A reaction is at equilibrium at $$100{\,^ \circ }C$$ and the enthalpy change for the reaction is $$42.6\,kJ\,mo{l^{ - 1}}.$$ What will be the value of $$\Delta S$$ in $$J\,{K^{ - 1}}\,mo{l^{ - 1}}?$$
A.
120
B.
426.2
C.
373.1
D.
114.2
Answer :
114.2
Solution :
$$\eqalign{
& \Delta G = \Delta H - T\Delta S \cr
& {\text{At equilibrium}},\Delta G = 0,\Delta H = T\Delta S \cr} $$
$$\Delta S = \frac{{\Delta H}}{T} = \frac{{42600\,J\,mo{l^{ - 1}}}}{{373\,K}}$$ $$ = 114.2\,J\,{K^{ - 1}}\,mo{l^{ - 1}}$$
Releted MCQ Question on Physical Chemistry >> Chemical Thermodynamics
Releted Question 1
The difference between heats of reaction at constant pressure and constant volume for the reaction : $$2{C_6}{H_6}\left( l \right) + 15{O_{2\left( g \right)}} \to $$ $$12C{O_2}\left( g \right) + 6{H_2}O\left( l \right)$$ at $${25^ \circ }C$$ in $$kJ$$ is
$${\text{The}}\,\Delta H_f^0\,{\text{for}}\,C{O_2}\left( g \right),\,CO\left( g \right)\,$$ and $${H_2}O\left( g \right)$$ are $$-393.5,$$ $$-110.5$$ and $$ - 241.8\,kJ\,mo{l^{ - 1}}$$ respectively. The standard enthalpy change ( in $$kJ$$ ) for the reaction $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$