Question

At same temperature $N_2{O}_4$ is dissociated to $40\%$ & $50\%$ at total pressure $P_1$ & $P_2$ atm respectively in $N{O}_2$. Then the ratio of $P_1$ & $P_2$ is:

• A. $\frac{4}{5}$
• B. $\frac{7}{4}$
• C. $\frac{4}{7}$
• D. None of these

Answer 1

The correct option is B $\frac{7}{4}$

$\begin{array}{ccc}& \begin{array}{cc}{N}_2{O}_4& =28+64=\underset{–}{92}\\ {\alpha }_1=& .4,{\alpha }_2=.5\\ P_1& =P_2=\\ N_2{O}_4& =2{NO}_2\end{array}& \\ & 1& 0\\ & 1-\alpha & \alpha \\ & \frac{1-\alpha }{1+\alpha }p\frac{2\alpha }{1+\alpha }r& \end{array}$

$\begin{array}{c}Kp={\left(\frac{2\alpha }{1+\alpha }\right)}^2\times \frac{1+\alpha }{(1-\alpha )}{p}_1\\ Kp=\frac{4{\alpha }^2}{1-{\alpha }^2}{P}_0\text{Gere}\end{array}$

$\begin{array}{cc}& \begin{array}{c}\text{Kp of any ran does not change}\\ \text{when rxn made is no change}\end{array}\\ & K_p=\frac{4{\alpha }_1^2}{1-{\alpha }_1^2}{P}_1\end{array}$

$\begin{array}{cc}{K}_P& =\frac{4{\alpha }_2}{{(1-{\alpha }_2^2)}^8}{P}_2\\ \frac{{\alpha }_1{P}_1}{1-{\alpha }_1^2}& =\frac{{\alpha }_2}{1-{\alpha }_2^2}{P}_2\\ \frac{P_1}{P_2}& =\frac{1-{\alpha }_1^2}{1-{\alpha }_2^2}\times \frac{{\alpha }_2^2}{{\alpha }_1}\\ & =\frac{1-(0.4{)}^2}{1-(0.5{)}^2}\times \frac{(0.5{)}^2}{(0.4{)}^2}\\ \frac{P_1}{P_2}& =\frac{7}{4}\end{array}$

Correct Answer =B