At what temperature liquid water will be in equilibrium with water vapour ?
$$\Delta {H_{{\text{vap}}}} = 40.73\,kJ\,mo{l^{ - 1}},$$ $$\Delta {S_{{\text{vap}}}} = 0.109\,kJ\,{K^{ - 1}}\,mo{l^{ - 1}}$$
A.
282.4$$\,K$$
B.
373.6$$\,K$$
C.
100$$\,K$$
D.
400$$\,K$$
Answer :
373.6$$\,K$$
Solution :
$${\text{At equilibrium}}\,\,\Delta G = 0\,\,$$ $${\text{for}}\,\,\Delta G = \Delta H - T\Delta S$$
$${\text{or}}\,\,T = \frac{{\Delta H}}{{\Delta S}} = \frac{{40.73}}{{0.109}} = 373.6\,K$$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$