Question

Partial pressure of A,B,C and D on the basis of gaseous system at equilibrium \[\mathbf{A}+\mathbf{2B}\rightleftharpoons \mathbf{C}+\mathbf{3D}~\] are \[\mathbf{A}=\mathbf{0}.\mathbf{20};\mathbf{B}=\mathbf{0}.\mathbf{10};\mathbf{C}=\mathbf{0}.\mathbf{30}~\mathbf{and}~\mathbf{D}=\mathbf{0}.\mathbf{50atm}\].
The numerical value of equilibrium constant is-
(A) 11.25
(B) 18.75
C. 5
D. 3.75

Answer 1

Hint: To solve this question we have to know about equilibrium constant. In an equilibrium reaction the rate constant of the reaction is given as the ratio of partial pressure of product to the partial pressure of reactant. Any coefficient of the reactant or product is used as the power of the partial pressure.

Formula Used: The formula used in this case is given as:
\[{{K}_{eq}}=\dfrac{{{P}_{C}}P_{D}^{3}}{{{P}_{A}}P_{B}^{2}}\]

Complete Step by Step Solution:
The given reaction is \[\mathbf{A}+\mathbf{2B}\rightleftharpoons \mathbf{C}+\mathbf{3D}~\].

Here one mole of reactant $A$ reacts with two moles of reactant $B$ to produce one mole of product $C$and three moles of product $D$.
Here the given partial pressure of reactant $A$ is $0.20atm$.
Here the given partial pressure of reactant $B$ is $0.10atm$.
Here the given partial pressure of product $C$ is $0.30atm$.
Here the given partial pressure of product D is $0.50atm$.

The rate constant of the above equilibrium reaction is given as follows-
\[{{K}_{eq}}=\dfrac{{{P}_{C}}P_{D}^{3}}{{{P}_{A}}P_{B}^{2}}\]
Putting the values of the partial pressure of the reactants and products in the above equation we get:
\[{{K}_{eq}}=\dfrac{0.3\times 0.5\times 0.5\times 0.5}{0.2\times 0.1\times 0.1}\]
\[{{K}_{eq}}=18.75\]
Thus the value of the equilibrium constant of the above reaction or \[Keq\]is $18.75$.
Thus the correct option is B.

Note: An equilibrium reaction is that reaction where there is an equilibrium between the reactants and products of the reaction. Here the reaction can proceed in both the forward and backward directions that means reactants react to give product and again product can also break down into the corresponding reactants.