Question

The dissociation equilibrium of gas \[A{B_2}\] can be represented as:
\[2A{B_2}(g) \rightleftharpoons 2AB(g) + {B_2}(g)\]
The degree of dissociation is x, and it is small compared to 1. The expression for relating the degree of dissociation x with equilibrium constant \[{K_P}\], and the total pressure p is:
(A) \[(2{K_P}/p)\]
(B) \[{(2{K_P}/p)^{\dfrac{1}{3}}}\]
(C) \[{(2{K_P}/p)^{\dfrac{1}{2}}}\]
(D) \[({K_P}/p)\]

Answer 1

Hint: The equilibrium constant expression is
\[{K_P} = \dfrac{{{{({P_{AB}})}^2} \times ({P_{{B_2}}})}}{{{{({P_{A{B_2}}})}^2}}}\]
The total pressure of the system is p. Since the degree of dissociation is small when compared to 1, 1−x can be approximated to 1.

Complete answer:
Let P be the initial pressure of \[A{B_2}\].
The equilibrium pressures of \[A{B_2}\], \[AB\]and \[{B_2}\] are P(1-x), xP and 0.5xP respectively.
The equilibrium constant expression is
\[{K_P} = \dfrac{{{{({P_{AB}})}^2} \times ({P_{{B_2}}})}}{{{{({P_{A{B_2}}})}^2}}}\]
\[{K_P} = \dfrac{{{{(xP)}^2} \times (0.5xP)}}{{{{[P(1 - x)]}^2}}}\]
\[{K_P} = P\dfrac{{0.5{x^3}}}{{{{(1 - x)}^2}}}\]
Since the degree of dissociation is small when compared to 1,1−x can be approximated to 1.
\[{K_P} = P \times 0.5{x^3}\]……. (1)
The total pressure is
\[P(1 - x) + Px + 0.5Px = P(1 + 0.5x) = p\]
\[p = \dfrac{p}{{1 + 0.5x}}\]……… (2)
Substituting (2) in (1)
\[{K_P} = \dfrac{p}{{1 + 0.5x}} \times 0.5{x^3}\]
Since the degree of dissociation is small compared to 1,1+0.5x can be approximate to 1.
\[{K_P} = p \times 0.5{x^3}\]
\[x = {\left( {\dfrac{{2{K_P}}}{p}} \right)^{\dfrac{1}{3}}}\]
Hence, the expression for relating the degree of dissociation (x) with equilibrium constant \[{K_P}\] and the total pressure is \[{\left( {\dfrac{{2{K_P}}}{p}} \right)^{\dfrac{1}{3}}}\]

Therefore, the correct answer is option(B).

Note:
Always look into the approximations whenever mentioned in the question. The equilibrium constant is constant for a reaction unless stated otherwise. The expression for the equilibrium constants is derived from thermodynamics, so technically each partial pressure term must be divided by its standard state value before putting it up in the formula. This makes the constant as unitless.