Question

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

Answer 1

Hint: 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 ;$A{B_{(g)}} \rightleftharpoons {A_{(g)}} + {B_{(g)}}$
\[
  \begin{array}{*{20}{c}}
  {{\text{At initial}}}&\;\;\;\;{{\text{ }}\;\;\;1}&{{\text{ }} - }&{{\text{ }} - }
\end{array} \\
  \begin{array}{*{20}{c}}
  {{\text{At equilibrium}}}&{1 - \alpha }&\alpha &\alpha
\end{array} \\
 \]
Then, the number of moles in the reaction is $1 + \alpha $.
$\alpha $ - degree of dissociation
Now, the total pressure of the reaction AB will be
${P_{A + B}} = \dfrac{{1 - \alpha }}{{1 + \alpha }}P$
Where,
${P_A} = \dfrac{\alpha }{{1 + \alpha }}P$ and ${P_B} = \dfrac{\alpha }{{1 + \alpha }}P$
So, ${K_P} = \dfrac{{{\alpha ^2}P}}{{1 - {\alpha ^2}}}$
Therefore, using this formula we will find the dissociation percentage of AB in the above reaction.
As we know that, Pressure = 1 atm = ${\alpha ^2}$
So, $\alpha = \sqrt K $
Here, we know that dissociation equilibrium K = $2.56 \times {10^{ - 2}}atm$.
Now, we will place the value of K in the above formula we get,
$ \Rightarrow \alpha = \sqrt {2.56 \times {{10}^{ - 2}}} $
On simplifying we get,
$ \Rightarrow \alpha = 0.16$
To calculate the percentage of dissociation equilibrium we will multiply with 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.