Question

The degree of dissociation of $N_2{O}_4$ into $N{O}_2$ at one atmosphere and ${40}^{o}C$ is $0.310$. If its $K_p$ at ${40}^{o}C$ is $X$ and the degree of dissociation (percentage) at $10atm$ pressure at same temperature is $Y$. Then $10X+100Y$ is:

Answer 1

$N_2{O}_4\rightleftharpoons 2N{O}_2$
Initial $10$
Final $1-0.310.62$
Total no. of moles$=1.31㏖$
$K=\frac{{\left(\frac{0.62}{1.31}\right)}^2}{\left(\frac{0.69}{1.31}\right)}=0.425atm$
At $10atm\frac{{\left(\frac{2\alpha }{1-\alpha }10\right)}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)10}=0.425$
$\Rightarrow \frac{\alpha +{\alpha }^2}{1-{\alpha }^2}=0.0425$
$\Rightarrow 94.12{\alpha }^2=1-{\alpha }^2$
$\alpha =0.1025$

$10X+100Y=10\times 0.425+100\times 0.1025=145$