Chemical EquilibriumHard
Question
For the equilibrium A(g) $\rightleftharpoons$nB(g), the equilibrium constant KP is related with the degree of dissociation α, and the total pressure of gases at equilibrium P is
Options
A.$\frac{(n\alpha)^{n}.P^{n - 1}}{(1 - \alpha).\left\lbrack 1 + (n - 1)\alpha \right\rbrack^{n - 1}}$
B.$\frac{(n\alpha)^{n - 1}.P^{n - 1}}{(1 - \alpha).\left\lbrack 1 + (n - 1)\alpha \right\rbrack^{n - 1}}$
C.$\frac{(\alpha)^{n - 1}.P^{n - 1}}{(1 - \alpha)^{n - 1}.\left\lbrack 1 + (n - 1)\alpha \right\rbrack}$
D.$\frac{n\alpha.P^{n - 1}}{(1 - \alpha)^{n - 1}.\left\lbrack 1 + (n - 1)\alpha \right\rbrack}$
Solution
A(g) $\rightleftharpoons$ nB(g)
Initial mole 1(say) 0
Equilibrium mole 1 –$\alpha$ n$\alpha$
Total moles = 1–$\alpha$ + n$\alpha$ = 1 + $\alpha$ (n – 1)
Now, $K_{P} = \frac{P_{B}^{n}}{P_{A}} = \frac{\left\lbrack \frac{n\alpha}{1 + \alpha(n - 1)}.P \right\rbrack^{n}}{\left\lbrack \frac{1 - \alpha}{1 + \alpha(n - 1)}.P \right\rbrack} = \frac{(n\alpha)^{n}.P^{n - 1}}{(1 - \alpha).\left\lbrack 1 + \alpha(n - 1) \right\rbrack^{n - 1}}$
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