Chemical EquilibriumHard
Question
For dissociation of a gas N2O5 as N2O5(g) $\rightleftharpoons$2NO2(g) + ½ O2(g). The reaction is performed at constant temperature and volume. If D is the vapour density of equilibrium mixture, Po is initial pressure of N2O5(g) and M is molecular mass of N2O5, then the correct information(s) at the equilibrium is/are
Options
A.the total pressure of gases at equilibrium is $\frac{P_{o}.M}{2D}$.
B.the degree of dissociation of N2O5(g) is $\frac{M - 2D}{3D}$.
C.the partial pressure of N2O5(g) at equilibrium is $\frac{(5D - M).P_{o}}{3D}$.
D.the partial pressure of O2(g) at equilibrium is $\frac{(M - 2D).P_{o}}{3D}$.
Solution
N2O5(g) $\rightleftharpoons$ 2NO2(g) + ½ O (g)
Initial partial pressure P0 0 0
Equ. partial pressure P0(1 – $\alpha$) 2$\alpha$P0 $\frac{\alpha P_{o}}{2}$
and $\alpha = \frac{M - 2D}{\left( \frac{5}{2} - 1 \right)2D} = \frac{M - 2D}{3D}$
Total equilibrium pressure = $P_{0}(1 - \alpha) + 2\alpha P_{0} + \frac{\alpha P_{0}}{2} = P_{0}\left( 1 + \frac{3\alpha}{2} \right)$
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