Question

The degree of dissociation of SO3 is $\alpha$ at equilibrium pressure P0 .

Kp for 2SO3(g) $\rightleftharpoons$ 2SO2(g) + O2(g) is

• A. $\lbrack(P_{0}\alpha^{3})\text{/2}(\text{1 - }\alpha)^{3}\rbrack$
• B. $\left\lbrack \left( P_{0}\alpha^{3} \right)/(2 + \alpha)(1 - \alpha)^{2} \right\rbrack$
• C. $\lbrack(P_{0}\alpha^{2})\text{/2}(\text{1 - }\alpha)^{2}\rbrack$
• D. None of these

Answer 1

Answer: (B)

$\left\lbrack \left( P_{0}\alpha^{3} \right)/(2 + \alpha)(1 - \alpha)^{2} \right\rbrack$

Answer 2

2SO3(g) $\rightleftharpoons$ 2SO2 (g) + O2(g)

t=0 a 0 0

t=teq. $a(1 - \alpha)$ a$\alpha$ $a\left( \frac{\alpha}{2} \right)$

Total mole at eq. $= a\left( 1 + \frac{\alpha}{2} \right)$

$P_{SO_{3}} = \left( \frac{1 - \alpha}{1 + (\alpha/2)} \right)P_{0}\left\lbrack \frac{(1 - \alpha)}{2 + \alpha} \right\rbrack \times P_{0};P_{SO_{2}} = \left( \frac{\alpha}{1 + (\alpha/2)} \right)P_{0} = \left( \frac{2\alpha}{2 + \alpha} \right) \times P_{0}PO_{2} = \left( \frac{\alpha/2}{1 + (\alpha/2)} \right)P_{0}$

$K_{P} = \frac{\frac{4\alpha^{2}\left( P^{0} \right)^{2}}{(2 + \alpha)^{2}} \times \left( \frac{\alpha}{2 + \alpha} \right) \times P^{0}}{\frac{4(1 - \alpha)^{2}}{\lbrack 2 + \alpha\rbrack^{2}} \times \left( P_{0} \right)^{2}} = \left\lbrack \frac{\alpha^{3}P^{0}}{(2 + \alpha)(1 - \alpha)^{2}} \right\rbrack$.