Question

The dissociation equilibrium of a gas \[A{B_2}\] can be represented as:
$2A{B_2}\left( g \right) \rightleftarrows 2AB\left( g \right) + {B_2}\left( g \right)$
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 the total pressure $p$ is:
(i) $\dfrac{{2{K_p}}}{p}$
(ii) ${\left( {\dfrac{{2{K_p}}}{p}} \right)^{\dfrac{1}{3}}}$
(iii) ${\left( {\dfrac{{2{K_p}}}{p}} \right)^{\dfrac{1}{2}}}$
(iv) $\dfrac{{{K_p}}}{p}$

Answer 1

Hint:The equilibrium constant of pressure $\left( {{K_p}} \right)$ is the equilibrium constant calculated from the partial pressures of a reaction equation. For a chemical reaction, the equilibrium constant can be defined as the ratio between the amount of reactants and the amount of products and is used to determine chemical behaviour.

Complete step-by-step answer:The equilibrium constant of a chemical reaction $($usually denoted by the symbol $K\,)$ provides insight into the relationship between the products and reactants when a chemical reaction reaches equilibrium. The equilibrium constant of pressure $\left( {{K_p}} \right)$ of a chemical reaction at equilibrium can be defined as the ratio of the partial pressure of products to the partial pressure of the reactants, each raised to their respective stoichiometric coefficients.
The dissociation equilibrium of a gas \[A{B_2}\] is given as:
$2A{B_2}\left( g \right) \rightleftarrows 2AB\left( g \right) + {B_2}\left( g \right).......\left( 1 \right)$.
Let ${P^ \circ }$be the initial pressure of \[A{B_2}\].
Therefore at time $t = 0$, the partial pressures of $A{B_2},\,AB$and ${B_2}$are ${P^ \circ },\,0,\,0$ respectively.
The degree of dissociation is $x$.
Therefore at time $t = t$, the equilibrium pressures of $A{B_2},\,AB$and ${B_2}$are ${P^ \circ }\left( {1 - x} \right),\,x{P^ \circ },\,0.5x{P^ \circ }$ respectively.
Hence, the equilibrium constant of pressure $\left( {{K_p}} \right)$ for the equation $\left( 1 \right)$is:
${K_p}\, = \,\dfrac{{{{\left( {{P_{AB}}} \right)}^2}\left( {{P_{{B_2}}}} \right)}}{{{{\left( {{P_{A{B_2}}}} \right)}^2}}}\, = \,\dfrac{{{{\left( {x{P^ \circ }} \right)}^2} \times \left( {0.5x{P^ \circ }} \right)}}{{{{\left[ {{P^ \circ }\left( {1 - x} \right)} \right]}^2}}}\, = \,{P^ \circ } \times \dfrac{{0.5{x^3}}}{{{{\left( {1 - x} \right)}^2}}}$
Since, $x < < < 1$ therefore $\left( {1 - x} \right) \sim 1$.
$ \Rightarrow \,{K_p}\, = \,{P^ \circ } \times 0.5{x^3}........\left( 2 \right)$
Now, applying Dalton’s Law of Partial Pressure we know that,
Total Pressure $\left( p \right)\, = \,{P^ \circ }\left( {1 - x} \right) + x{P^ \circ } + 0.5x{P^ \circ }$
$ \Rightarrow \,p\, = \,{P^ \circ } - x{P^ \circ } + x{P^ \circ } + 0.5x{P^ \circ }$
$ \Rightarrow \,p\, = \,{P^ \circ } + 0.5x{P^ \circ }$
$ \Rightarrow \,p\, = \,\left( {1 + 0.5x} \right){P^ \circ }$
$ \Rightarrow \,{P^ \circ }\, = \,\dfrac{p}{{1 + 0.5x}}$
Putting the value of $p$in equation $\left( 2 \right)$ we get,
${K_p}\, = \,\dfrac{p}{{1 + 0.5x}} \times 0.5{x^3}$
Since, $x < < < 1$ therefore $\left( {1 + 0.5x} \right) \sim 1$.
\[ \Rightarrow \,{K_p}\, = \,p \times 0.5{x^3}\]
$ \Rightarrow \,{x^3}\, = \,\dfrac{{{K_p}}}{{0.5p}}\, = \,\dfrac{{2{K_p}}}{p}$
$ \Rightarrow \,x\, = \,{\left( {\dfrac{{2{K_p}}}{p}} \right)^{\dfrac{1}{3}}}$

Hence the correct answer is (ii) ${\left( {\dfrac{{2{K_p}}}{p}} \right)^{\dfrac{1}{3}}}$.

Note: It is important to note that there are several different types of equilibrium constants that provide relationships between the products and the reactants of equilibrium reactions in terms of different units so you must make sure which one is being asked in the given question otherwise it will lead to error.