Question

What is free energy change $\left(\Delta G\right)$ when $1.0$ mole of water at ${100}^{o}C$ and $1$ atm pressure is converted into steam at ${100}^{o}C$ and $2atm$ pressure?

• A. $0cal$
• B. $540cal$
• C. $515.4cal$
• D. $Noneofthese$

Answer 1

The correct option is A $0cal$

• Since, the reaction is being carried out at constant temperature $\left({100}^{0}C\right)$, therefore the reaction is in thermal equilibrium. At equilibrium the value of $\Delta G$ is zero.
  
• This can also be proved mathematically

$\Delta G=\Delta G^0+RTlnQ$

At equilibrium, $\Delta G^0=-RTln{K}_{eq}$ and $K_{eq}=Q$

(where $Q$ is the ratio of initial activity of products to reactants, and $K_{eq}$ is the ratio of concentrations of products to reactants at equilibrium)

Thus, $\Delta G=-RTlnQ+RTlnQ=0$