Question

$N_2{O}_4$ is dissociated to 33% and 50% at total pressure $P_1$ and $P_2$ atm respectively. The ratio of $\frac{P_1}{P_2}$is:

• A. 7/4
• B. 7/3
• C. 8/3
• D. 8/5

Answer 1

The correct option is C

8/3

$\begin{array}{ccc}& N_2{O}_4\left(g\right)& \leftrightharpoons 2N{O}_2\left(g\right)\\ Intialno.ofmoles& 1& 0\\ No.ofmolesatEqb& (1-x)& 2x\end{array}$

Total moles $N_{total}$ = (1 - x) + 2x = (1 - x)

Partial pressure of $N_2{O}_4$(g) & $N{O}_2$ (g) are

$P_{N_2{O}_4}$ = $\frac{n_{N_2{O}_4}}{n_{total}}$ $\times$ P = $\frac{(1-x)}{(1+x)}$ $\times$ P ; $P_{N{O}_2}$= $\frac{n_{N{O}_2}}{n_{total}}$ $\times$ P = $\frac{2x}{(1+x)}$ $\times$P

From the above equation Let $X_1$ = 33% = 0.33 & $x_2$ = 50% = 0.5

$K_p$ = $\frac{{x_1}^2\times P}{(1-{x_1}^2)}$ = $\frac{{x_2)}^2\times P}{(1-{x_2}^2)}$ ; $\frac{{\left(0.33\right)}^2\times P_1}{(1-{\left(0.33\right)}^2)}$ = $\frac{{\left(0.5\right)}^2\times P_2}{(1-{\left(0.5\right)}^2)}$ = $\frac{P_1}{P_2}$ = $\frac{8}{2}$