
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
162k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main Mock Test Series Class 12 Chemistry for FREE

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Classification of Drugs

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

Other Pages
NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Solutions Class 12 Notes: CBSE Chemistry Chapter 1

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

Electrochemistry Class 12 Notes: CBSE Chemistry Chapter 2
