Question

For the equilibrium at $298$ K; $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right);G_{N_2{O}_4}^{\ominus }=100kJmo{l}^{-1}$ and $G_{N{O}_2}^{\ominus }=50kJmo{l}^{-1}$. If 5 mol of $N_2{O}_4$ and 2 moles of $N{O}_2$ are taken initially in one litre container than which statement are correct

• A. reaction proceeds in forward direction
• B. $K_c=1$
• C. $\Delta G=-0.55kJ,\Delta G^{\ominus }=0$
• D. At equilibrium $\left[N_2{O}_4\right]=4.84M$ and $\left[N{O}_2\right]=0.212M$

Answer 1

Answer: (A), (B), (C)

The correct options are
A reaction proceeds in forward direction
B $K_c=1$
C $\Delta G=-0.55kJ,\Delta G^{\ominus }=0$

$\Delta G=\Delta G^{\ominus }+2.303RTlogQ$

$\Delta G^{\ominus }=2\times G_{N{O}_2}^{\ominus }-G_{N_2{O}_4}^{\ominus }=2\times 50-100=0$

$\therefore \Delta G=0+2.303\times 8.314\times {10}^{-3}\times 298log\frac{22}{5}=0-0.55kJ$

$\therefore \Delta G=-0.55kJ$, i.e, reaction proceeds in forward direction

Also $\Delta G^{\ominus }=0=2.303RTlogK\therefore K=1$

Now, $\underset{\underset{5-x}{5}}{N_2{O}_4}=\underset{\underset{2+2x}{2}}{2N{O}_2}$

$\therefore K_p=\frac{\left(P_{N{O}_2}\right)}{\left(P_{N_2{O}_4}\right)}=1=\frac{(2+2x{)}^2}{5-x}$ or $x=0.106$

So, $\left[N_2{O}_4\right]=5-x=4.894M,\left[N{O}_2\right]=2+2x=2.12M$

Hence, options A, B and C are correct.