Chemistry - Thermodynamics - II Question with Solution | TestHub

$${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2N{\mathrm{O}}_2$$Initial moles:

$$\begin{array}{lcc}Initialmoles& 1& 0\\ Ateqmmole& 1-\alpha & 2\alpha \end{array}$$

The equilibrium constant $K_P$ is given by:

$$K_P=\frac{(2\alpha {)}^{2}P}{(1-{\alpha }^2)}=\frac{4(0.5{)}^2\times 1}{1-(0.5{)}^2}=\frac{4}{3}$$

The standard Gibbs free energy change $\Delta G^{\circ }$ is calculated as:

$$\Delta G^{\circ }=-RT\ln K_P$$Substituting the values:

$$=-2\times 300\ln \left(\frac{4}{3}\right)$$

$$=-2\times 300\times 0.3=-180cal$$

Therefore, the absolute value of the standard Gibbs free energy change is:

$$|\Delta G^{\circ }|=180cal$$