Question

If in the reaction $N_2{O}_4\left(g\right)\leftrightharpoons 2N{O}_2\left(g\right),\alpha$ is the dissociation constant at equilibrium, then the number of moles at equilibrium will be:

Note: Initially there is only 1 mol.

• A. $3$
• B. $1$
• C. ${(1-\alpha )}^2$
• D. $(1+\alpha )$

Answer 1

The correct option is D $(1+\alpha )$

Given the reaction is as follows:

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

Here, $\alpha$ is the degree of dissociation at equilibrium.

Suppose, initially $1$ mole of $N_2{O}_4$ and $0$ moles of $N{O}_2$ are present.

If $\alpha$ is the degree of dissociation then, at equilibrium $(1-\alpha )$ moles of $N_2{O}_4$ and $2\alpha$ moles of $N{O}_2$ are present.

Therefore, the total number of moles present at equilibrium are $(1-\alpha )+2\alpha =(1+\alpha )$.