Chemical And Ionic Equilibrium Result Question 43
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 K and the total pressure was found to be 20 bar. Given that
$\Delta G_f^{\circ}\left(\mathrm{N}_2 \mathrm{O}_4\right)=-9.16 \mathrm{~kJ}, \Delta G_f^{\circ}\left(\mathrm{NO}_2\right)=+33.15 \mathrm{~kJ}$
(i) Find $\Delta G$ of the reaction.
(ii) The direction in which the reaction shifts to reach equilibrium.
(b) A graph is plotted for a real gas which follows van der Waals’ equation with $V_m$ taken on $Y$-axis and $p$ on $X$-axis. Find the intercept of the line where $V_m$ is molar volume.
(2004, 4M)
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Solution:
- (a) $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$
$\Delta G^{\circ}=2 \Delta G_f^{\circ}\left(\mathrm{NO}_2\right)-\Delta G_f^{\circ}\left(\mathrm{N}_2 \mathrm{O}4\right)=\Delta G^{\circ}{\text{rxn}}$
Also $\quad \Delta G^{\circ}=-R T \ln K=0, \quad K=1$
Let the reaction shift in the forward direction.
$\quad\quad \mathrm{N}_2 \mathrm{O}_4(g) \quad \rightleftharpoons\quad 2 \mathrm{NO}_2(g) \quad \text { Total } $
$\quad\quad\quad5-x \quad \quad \quad\quad 5+2 x \quad \quad 10+x $
$p_i: \dfrac{5-x}{10+x} \times 20 \quad\quad\quad\quad \dfrac{5+2 x}{10+x} \times 20 $
$\Rightarrow K=\dfrac{(5+2 x)^2}{(10+x)^2} \times \dfrac{10+x}{5-x} \times 20=1 $
$\Rightarrow 81 x^2+405 x+450=0 $
$x=-1.66 \text { and }-3.33$
Both values of $x$ indicate that reaction actually proceeds in backward direction.
$\begin{aligned} & \left(p+\dfrac{a}{V_m^2}\right)\left(V_m-b\right)=R T \\ & \left(p+\dfrac{a}{V_m^2}\right)\left(V_m-b\right)=R T\end{aligned}$
$\Rightarrow\left[\left(p V^2\right) p+a p^2\right][(p V)-b]=p(p V)^2 R T$
$\Rightarrow p. [pV^2 + ap](pV-bp)=(pV^2)RT $
But $p=0$
$\text { Intercept }=R T \Rightarrow(p V)^2=(p V)^2 R T$
BETA
