The degree of dissociation of SO3 is α at equilibrium pressure P0.Kp for 2SO3(g)⇌2SO2(g)+O2(g) is:

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The degree of dissociation of ${SO}_{3}$ is $α$ at equilibrium pressure $P_{0}$ .

$K_{p} for 2 {SO}_{3} (g) ⇌ 2 {SO}_{2} (g) + O_{2} (g)$ is:

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$[\frac{(P_{0} α^{3})}{2 (1 - α)^{2}}]$

$[\frac{(P_{0} α^{3})}{(2 + α) (1 - α)^{2}}]$

$[\frac{(P_{0} α^{3})}{2 (1 - α)^{3}}]$

None of these

answer is D.

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Detailed Solution

In the following reaction we have,

$2 {SO}_{3} ⇌ 2 {SO}_{2} + O_{2}$

$a (1 - α) a α \frac{a α}{2}$

$\begin{array}{l} n_{T} = a (1 + \frac{α}{2}) \\ X_{{SO}_{3}} = \frac{1 - α}{1 + \frac{α}{2}}; X_{{SO}_{2}} = \frac{α}{1 + \frac{α}{2}} \\ X_{O_{2}} = \frac{\frac{α}{2}}{1 + \frac{α}{2}} \end{array}$

In the following reaction we have,

$K_{P} = \frac{{(\frac{α}{1 + \frac{α}{2}})}^{2} P_{0}^{2} \times (\frac{\frac{α}{2}}{1 + \frac{α}{2}}) \times P_{0}}{(\frac{1 - α}{1 + \frac{α}{2}}) \times P_{0}^{2}}$