What is the free energy change $$'\Delta G'$$ when $$1.0\,mole$$ of water at $${100^ \circ }C$$ and $$1\,atm$$ pressure is converted into steam at $${100^ \circ }C$$ and $$1\,atm.$$ pressure
A.
$$540\,cal$$
B.
$$ - 9800\,cal$$
C.
$$9800\,cal$$
D.
$$0\,cal$$
Answer :
$$0\,cal$$
Solution :
Condition of equilibrium, hence $$\Delta G = 0$$
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 $$C{O_2}\left( g \right) + {H_2}\left( g \right) \to CO\left( g \right) + {H_2}O\left( g \right)\,{\text{is}}$$