Chemical and Ionic Equilibrium - Result Question 43
####43. (a) In the following equilibrium $N _2 O _4(g) \rightleftharpoons 2 NO _2(g)$ when 5 moles of each are taken, the temperature is kept at 298 $K$ the total pressure was found to be 20 bar. Given that
$$ \Delta G _f^{\circ}\left(N _2 O _4\right)=100 kJ, \Delta G _f^{\circ}\left(NO _2\right)=50 kJ $$
(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. $\quad(2004,4 M)$
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Solution:
- (a) $N _2 O _4(g) \rightleftharpoons 2 NO _2(g)$
$$ \Delta G^{\circ}=2 \Delta G _f^{\circ}\left(NO _2\right)-\Delta G _f^{\circ}\left(N _2 O _4\right)=0 $$
Also $\quad \Delta G^{\circ}=-R T \ln K=0, \quad K=1$
Let the reaction shifts in forward direction.
$$ \begin{array}{rcc} & N _2 O _4(g) \rightleftharpoons 2 NO _2(g) \quad \text { Total } \ & 5-x & 5+2 x \quad 10+x \ p _i: & \frac{5-x}{10+x} \times 20 & \frac{5+2 x}{10+x} \times 20 \ \Rightarrow & K=\frac{(5+2 x)^{2}}{(10+x)^{2}} & \times \frac{10+x}{5-x} \times 20=1 \ \Rightarrow & 81 x^{2}+ & 405 x+450=0 \ & & x=-1.66 \text { and }-3.33 \end{array} $$
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 $=R T \Rightarrow(p V)^{3}=(p V)^{2} R T$
