Question

For the reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$, the value of $K_p$ is $1.7\times {10}^3$ at 500 K and $1.7\times {10}^4$ at 600 K. Which of the following is/are correct?

• A. The proportions of $N{O}_2$ in the equilibrium mixture is increased by decrease in pressure.
• B. The standard enthalpy change for the forward reaction is negative.
• C. Units of $K_p$ are $at{m}^{-1}$.
• D. At 500 K the degree of dissociation of $N_2{O}_4$ decreases by $50$% by increasing the pressure by $100$%.
• E. At $500$ K the degree of dissociation of $N_2{O}_4$ decreases by 50% by increasing the pressure by 200%.

Answer 1

Answer: (A)

The proportions of $N{O}_2$ in the equilibrium mixture is increased by decrease in pressure.

$\Delta n=2-1=1$
a. That is,with the decrease of pressure, reaction shifts towards right, i.e. proportions of $N{O}_2$ increases. Statement (a) is correct.

b. Value of K increases with increase of temperature and hence reaction is endothermic, i:e., $\Delta H=+ve$ Hence statement (b) is incorrect.

c. $K_p=\frac{[p_{N{O}_2}{]}^2}{\left[P_{N_2{O}_4}\right]}=\frac{at{m}^2}{atm}=atm$
Hence statement (c) is incorrect.

d. $\underset{\underset{1-\alpha }{1atm}}{N_2{O}_4\left(g\right)}\rightleftharpoons \underset{\underset{2\alpha }{0}}{2N{O}_2\left(g\right)}$
$K_p=\frac{(2\alpha {)}^2}{1-\alpha }=1.7\times {10}^3$ at 500 K
$\left[\right(1-\alpha \approx 1)$, since $\alpha$ is small]
$4{\alpha }^2=1.7\times {10}^3$
${\alpha }_2=\sqrt{\frac{1.7\times {10}^3}{4}}$
$\alpha =0.206\times {10}^2$
When the pressure was 1 atm, now let it be 2 atm (100% increase).
$\underset{\underset{2-\alpha }{2}}{N_2{O}_4\left(g\right)}\rightleftharpoons \underset{\underset{2\alpha }{0}}{2N{O}_2\left(g\right)}$
$K_p=\frac{(2\alpha {)}^2}{2-\alpha }=1.7\times {10}^2$ at 500 K
$[2-\alpha \approx 2$, since $\alpha$ is small]
$\frac{4{\alpha }^2}{2}=1.7\times {10}^3$
${\alpha }_2=\sqrt{\frac{1.7\times {10}^3}{2}}$
$\frac{{\alpha }_2}{{\alpha }_1}=\sqrt{\frac{1.7\times {10}^3\times 4}{2\times 1.7\times {10}^3}}=\sqrt{2}=1.4$
$\therefore$ When the pressure is increased 100% the decrease in $\alpha$ is 1.4 times which is not 50%. Hence statement is wrong.

e. $K_p$ at 600 K $=1.78\times {10}^4$
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$\Delta n=2-1=1$
Since by decrease of pressure reaction goes forward; i.e., more of $N_2{O}_4$ will dissociate. It means by decreasing pressure dissociation of $N_2{O}_4$ increases.
Hence, the statement is wrong.