What is the free energy change for the conversion of $$1\,mole$$ of water into steam at $$373.2\,K.$$ The heat of vaporization $$\left( {\Delta {H_v}} \right)$$ of water of $$373.2\,K$$ is $$9.1\,kcal\,mo{l^{ - 1}}.$$ The entropy change is $$25.5\,cal/mol\,\deg .$$
A.
$$ - 401.6\,cal/mol$$
B.
$$ - 416.6\,cal/mol$$
C.
$$516.5\,cal/mol$$
D.
$$ - 516.5\,cal/mol$$
Answer :
$$ - 416.6\,cal/mol$$
Solution :
$$\eqalign{
& \Delta G = \Delta H - T\Delta S \cr
& \therefore \,\,\Delta G = 9100 - 373.2 \times 25.5 \cr
& = - 416.6\,cal\,mo{l^{ - 1}} \cr} $$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$