Question

For the reaction :$P\rightleftharpoons Q+R$ .Initially 2 moles P was taken. Up to equilibrium $0.5$ moles of P was dissociated . What would be the degree of dissociation?

Answer 1

$P\leftrightharpoons Q+R$
initially , No of moles = $2$
at equilibrium $(2-0.5)$ $0.5$ $0.5$
$\alpha$ = degree of dissociation = $\frac{noofmolesdissociated}{noofmolespresentinitially}$
$\alpha =\frac{0.5}{2.0}=0.25$