Question

For which of the following reaction the degree of dissociation $\left(\alpha \right)$ and equilibrium constant $\left(K_P\right)$ are related as $K_P=\frac{4{\alpha }^{2}P}{(1-{\alpha }^2)}$ :-

• A. $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
• B. $H_2\left(g\right)+l_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$
• C. $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$
• D. $PC{I}_3\left(g\right)+C{I}_2\left(g\right)\rightleftharpoons PC{I}_5\left(g\right)$

Answer 1

The correct option is A $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$

$N_2{O}_4\left(g\right)=2N{O}_2\left(g\right)$

Initially: 1 0

Finally: 1-a 2a

Let х = mole fraction = number of moles/ total moles ; P= total pressure

Now, total moles = 1-a+2a = 1+a

We know, $Kp=(PN{O}_2{)}^2/P{N}_2{O}_4$

$=(хN{O}_2\times P{)}^2/(хN_2{O}_4\times P)$

$=\left(2aP/\right(1+a\left){)}^2/\right((1-a)P/(1+a))$

$=4a^{2}P/\left(\right(1+a\left)\right(1-a\left)\right)$

$=4a^{2}P/(1-a^2)$

Hence, option (1) $N_2{O}_4\left(g\right)=2N{O}_2\left(g\right)$ is correct.