Solveeit Logo
Login

Question

Question: In the reaction \(A{B_{(g)}} \rightleftharpoons {A_{(g)}} + {B_{(g)}}\) at \(30^\circ C\) , \({K_p}n...

In the reaction AB(g)⇌A(g)+B(g)A{B_{(g)}} \rightleftharpoons {A_{(g)}} + {B_{(g)}} at 30∘C30^\circ C , Kpn{K_p}n for the dissociation equilibrium is 2.56×10−2atm2.56 \times {10^{ - 2}}atm . If the total pressure at equilibrium is 1 atm, then the percentage dissociation of AB is?

Explanation

Solution

For solving this question, first we need to understand the meaning of dissociation reaction. We can say that a chemical reaction in which a given compound breaks apart into two or more components is a dissociation reaction. We can represent a dissociation reaction in the form:
AB →\to A + B.

Complete step by step answer:
We can write the given reaction as,
Chemical reaction ;AB(g)⇌A(g)+B(g)A{B_{(g)}} \rightleftharpoons {A_{(g)}} + {B_{(g)}}

\begin{array}{*{20}{c}} {{\text{At initial}}}&\;\;\;\;{{\text{ }}\;\;\;1}&{{\text{ }} - }&{{\text{ }} - } \end{array} \\\ \begin{array}{*{20}{c}} {{\text{At equilibrium}}}&{1 - \alpha }&\alpha &\alpha \end{array} \\\

Then, the number of moles in the reaction is 1+α1 + \alpha .
α\alpha - degree of dissociation
Now, the total pressure of the reaction AB will be
PA+B=1−α1+αP{P_{A + B}} = \dfrac{{1 - \alpha }}{{1 + \alpha }}P
Where,
PA=α1+αP{P_A} = \dfrac{\alpha }{{1 + \alpha }}P and PB=α1+αP{P_B} = \dfrac{\alpha }{{1 + \alpha }}P
So, KP=α2P1−α2{K_P} = \dfrac{{{\alpha ^2}P}}{{1 - {\alpha ^2}}}
Therefore, using this formula we will find the dissociation percentage of AB in the above reaction.
As we know that, Pressure = 1 atm = α2{\alpha ^2}
So, α=K\alpha = \sqrt K
Here, we know that dissociation equilibrium K = 2.56×10−2atm2.56 \times {10^{ - 2}}atm.
Now, we will place the value of K in the above formula we get,
⇒α=2.56×10−2\Rightarrow \alpha = \sqrt {2.56 \times {{10}^{ - 2}}}
On simplifying we get,
⇒α=0.16\Rightarrow \alpha = 0.16
To calculate the percentage of dissociation equilibrium we will multiply with 100.
α=16100×100\alpha = \dfrac{{16}}{{100}} \times 100 = 16%

Therefore, the percentage dissociation of AB is 16% in the reaction.

Note: After solving this question, we need to understand the concept of dissociation constant for better understanding. Dissociation constant is defined as a specific type of equilibrium constant that is responsible for measuring the propensity of a larger object to separate object reversibly into smaller components. For example, when a complex component falls apart into its component molecules, or when a salt splits up into its component ions. Then, the type of equilibrium constant found in the reaction is called dissociation constant.