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Degree of Dissociation and Kp
Concepts tested here
- Kp in terms of degree of dissociation
- Relating degree of dissociation to Kp
All Questions
2008 AIPMT 2 questions
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The dissociation equilibrium of a gas $AB_2$ can be represented as :
$2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$
The degree of dissociation is '$x$' and is small compared to 1. The expression relating the degree of dissociation ($x$) with equilibrium constant $K_p$ and total pressure $P$ is -$2AB_2(g) \rightleftharpoons 2AB(g) + B_2(g)$
Initial moles: 2, 0, 0
Moles at equilibrium: $2(1 - x)$, $2x$, $x$
Total amount of moles at equilibrium $= 2(1 - x) + 2x + x = 2 + x$
Partial pressures: $p_{AB_2} = \dfrac{2(1 - x)}{2 + x}P$, $p_{AB} = \dfrac{2x}{2 + x}P$, $p_{B_2} = \dfrac{x}{2 + x}P$
$K_p = \dfrac{(p_{AB})^2\,(p_{B_2})}{(p_{AB_2})^2}$
$K_p = \dfrac{\left(\dfrac{2x}{2 + x}P\right)^2 \left(\dfrac{x}{2 + x}P\right)}{\left(\dfrac{2(1 - x)}{2 + x}P\right)^2} = \dfrac{4x^3 P}{4(1 - x)^2 (2 + x)}$
As $x$ is small, $1 - x \approx 1$ and $2 + x \approx 2$:
$K_p = \dfrac{4x^3 P}{4 \times 2} = \dfrac{x^3 P}{2}$
$x^3 = \dfrac{2K_p}{P}$
$x = \left(\dfrac{2K_p}{P}\right)^{1/3}$ -
The values of $K_{p_1}$ and $K_{p_2}$ for the reactions
$X \rightleftharpoons Y + Z$ ....(1) and
$A \rightleftharpoons 2B$ ....(2)
are in ratio of 9 : 1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio :For $X \rightleftharpoons Y + Z$: initially 1, 0, 0; at equilibrium $1 - \alpha$, $\alpha$, $\alpha$.
Total number of moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha$
$K_{p_1} = \dfrac{p_Y \times p_Z}{p_X} = \dfrac{\left(\dfrac{\alpha}{1 + \alpha}P_1\right)\left(\dfrac{\alpha}{1 + \alpha}P_1\right)}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_1} = \dfrac{\alpha^2 P_1}{1 - \alpha^2}$
For $A \rightleftharpoons 2B$: initially 1, 0; at equilibrium $1 - \alpha$, $2\alpha$.
Total number of moles at equilibrium $= 1 - \alpha + 2\alpha = 1 + \alpha$
$K_{p_2} = \dfrac{(p_B)^2}{p_A} = \dfrac{\left(\dfrac{2\alpha}{1 + \alpha}P_2\right)^2}{\left(\dfrac{1 - \alpha}{1 + \alpha}\right)P_2} = \dfrac{4\alpha^2 P_2}{1 - \alpha^2}$
The degree of dissociation $\alpha$ is the same in both, so $\dfrac{K_{p_1}}{K_{p_2}} = \dfrac{P_1}{4P_2}$
$\dfrac{9}{1} = \dfrac{P_1}{4P_2}$
$\dfrac{P_1}{P_2} = \dfrac{36}{1}$, i.e. 36 : 1