Chemical and Ionic Equilibrium 1 Question 41
43. (a) In the following equilibrium $\mathrm{N}{2} \mathrm{O}{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$ when 5 moles of each are taken, the temperature is kept at 298 $\mathrm{K}$ the total pressure was found to be 20 bar. Given that
$$ \Delta G_{f}^{\circ}\left(\mathrm{N}{2} \mathrm{O}{4}\right)=100 \mathrm{~kJ}, \Delta G_{f}^{\circ}\left(\mathrm{NO}_{2}\right)=50 \mathrm{~kJ} $$
(i) Find $\Delta G$ of the reaction.
(ii) The direction of the reaction in which the equilibrium shifts.
(b) A graph is plotted for a real gas which follows van der Waals’ equation with $p V_{m}$ taken on $Y$-axis and $p$ on $X$-axis. Find the intercept of the line where $V_{m}$ is molar volume. $\quad(2004,4 \mathrm{M})$
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Solution:
- (a) $\mathrm{N}{2} \mathrm{O}{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$
$$ \Delta G^{\circ}=2 \Delta G_{f}^{\circ}\left(\mathrm{NO}{2}\right)-\Delta G{f}^{\circ}\left(\mathrm{N}{2} \mathrm{O}{4}\right)=0 $$
Also $\quad \Delta G^{\circ}=-R T \ln K=0, \quad K=1$
Let the reaction shifts in forward direction.
$$ \begin{array}{rcc} & \mathrm{N}{2} \mathrm{O}{4}(g) \rightleftharpoons 2 \mathrm{NO}{2}(g) \quad \text { Total } \ & 5-x & 5+2 x \quad 10+x \ p{i}: & \frac{5-x}{10+x} \times 20 & \frac{5+2 x}{10+x} \times 20 \ \Rightarrow & K=\frac{(5+2 x)^{2}}{(10+x)^{2}} & \times \frac{10+x}{5-x} \times 20=1 \ \Rightarrow & 81 x^{2}+ & 405 x+450=0 \ & & x=-1.66 \text { and }-3.33 \end{array} $$
Both values of $x$ indicates that reaction actually proceeds in backward direction.
(b) $\left(p+\frac{a}{V m^{2}}\right)\left(V_{m}-b\right)=R T$
$\left(p+\frac{a p^{2}}{(p V)^{2}}\right)\left(\frac{p V}{p}-b\right)=R T$
$\Rightarrow\left[\left(p V^{2}\right) p+a p^{2}\right][(p V)-b]=p(p V)^{2} R T$
$\Rightarrow p \cdot\left[p V^{2}+a p\right](p V-b p)=p\left(p V^{2}\right) R T$
But $p=0$
Intercept $=R T \Rightarrow(p V)^{3}=(p V)^{2} R T$
