
2 moles of \[PC{{l}_{5}}\] were heated in a closed vessel of 2 liters capacity. At equilibrium, $40%$ of \[PC{{l}_{5}}\] is dissociated into \[PC{{l}_{3}}\] and \[C{{l}_{2}}\] . The value of equilibrium constant is-
A. 0.266
B. 0.53
C. 2.66
D 5.3
Answer
161.4k+ 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 concentration of product to the concentration of reactant. Any coefficient of the reactant or product is used as the power of the concentration.
Formula Used: The formula used in this case is given as-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]
Complete Step by Step Solution:
The reaction can be written as\[PC{{l}_{5}}\left( g \right)\rightleftharpoons PC{{l}_{3}}\left( g \right)+C{{l}_{2}}(g)\].
Initial moles of \[PC{{l}_{5}}\]is 2 moles.
After $40%$ decomposition the moles of \[PC{{l}_{5}}\]left is given as
$ & 2-2\times \dfrac{40}{100} \\ $
$2-0.8=1.2$
Thus the amount of phosphorus trichloride and chlorine gas form is given as $0.8moles$.
Total volume of the vessel is 2 litres.
Thus the molar concentration of \[PC{{l}_{5}}\]is $\dfrac{1.2}{2}=0.6$.
The molar concentration of $PC{{l}_{3}}$ is $\dfrac{0.8}{2}=0.4$.
The molar concentration of $C{{l}_{2}}$ is $\dfrac{0.8}{2}=0.4$
The rate constant of the above equilibrium reaction is given as follows-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]
Putting the values of the concentration of the reactants and products in the above equation we get:
\[{{K}_{eq}}=\dfrac{0.4\times 0.4}{0.6}\]
\[{{K}_{eq}}=0.266\]
Thus the value of the equilibrium constant of the above reaction or \[Keq\] is $0.266$.
Thus the correct option is A.
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-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]
Complete Step by Step Solution:
The reaction can be written as\[PC{{l}_{5}}\left( g \right)\rightleftharpoons PC{{l}_{3}}\left( g \right)+C{{l}_{2}}(g)\].
Initial moles of \[PC{{l}_{5}}\]is 2 moles.
After $40%$ decomposition the moles of \[PC{{l}_{5}}\]left is given as
$ & 2-2\times \dfrac{40}{100} \\ $
$2-0.8=1.2$
Thus the amount of phosphorus trichloride and chlorine gas form is given as $0.8moles$.
Total volume of the vessel is 2 litres.
Thus the molar concentration of \[PC{{l}_{5}}\]is $\dfrac{1.2}{2}=0.6$.
The molar concentration of $PC{{l}_{3}}$ is $\dfrac{0.8}{2}=0.4$.
The molar concentration of $C{{l}_{2}}$ is $\dfrac{0.8}{2}=0.4$
The rate constant of the above equilibrium reaction is given as follows-
\[{{K}_{eq}}=\dfrac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}\]
Putting the values of the concentration of the reactants and products in the above equation we get:
\[{{K}_{eq}}=\dfrac{0.4\times 0.4}{0.6}\]
\[{{K}_{eq}}=0.266\]
Thus the value of the equilibrium constant of the above reaction or \[Keq\] is $0.266$.
Thus the correct option is A.
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 Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE 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

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

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

NCERT Solutions for Class 12 Chemistry Chapter 1 Solutions

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

NCERT Solutions for Class 12 Chemistry Chapter 2 Electrochemistry

NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Solutions Class 12 Notes: CBSE Chemistry Chapter 1
