
The unit of equilibrium constant $K$ for the reaction \[A+B\rightleftharpoons C~\] would be
A. $mol.lite{{r}^{-1}}$
B. $mo{{l}^{-1}}litre$
C. mol. litre
D. Dimensionless
Answer
162.6k+ views
Hint: To solve this question we have to know about equilibrium constant and unit of 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. The unit of concentration is given as $mol.lite{{r}^{-1}}$.
Complete Step by Step Answer:
The given reaction is \[A+B\rightleftharpoons C~\].
Here one mole of reactant $A$ reacts with two moles of reactant $B$ to produce one mole of product $C$. It is a not an unimolecular reaction as the change in number of moles is not zero. It is an equilibrium reaction as the product can also decompose to give back the reactants.
Thus the equilibrium constant is given as-
$K=\dfrac{[C]}{[A][B]}$
The concentration of reactant and product has the unit of $mol.lite{{r}^{-1}}$ .
$K=\dfrac{mol.lite{{r}^{-1}}}{{{(mol.lite{{r}^{-1}})}^{2}}}$
$K=mo{{l}^{(1-2)}}.lite{{r}^{(-1+2)}}$
$K=mo{{l}^{-1}}.litre$
Thus the unit of equilibrium constant is $mo{{l}^{-1}}.litre$.
Thus the correct option is B.
Note: Units of equilibrium constant K will depend on the equilibrium of the concentration of the number of reactants to the products. If the total number of moles of product is equal to the total number of moles of reactant then the equilibrium constant K has no units i.e. it is dimensionless.
Complete Step by Step Answer:
The given reaction is \[A+B\rightleftharpoons C~\].
Here one mole of reactant $A$ reacts with two moles of reactant $B$ to produce one mole of product $C$. It is a not an unimolecular reaction as the change in number of moles is not zero. It is an equilibrium reaction as the product can also decompose to give back the reactants.
Thus the equilibrium constant is given as-
$K=\dfrac{[C]}{[A][B]}$
The concentration of reactant and product has the unit of $mol.lite{{r}^{-1}}$ .
$K=\dfrac{mol.lite{{r}^{-1}}}{{{(mol.lite{{r}^{-1}})}^{2}}}$
$K=mo{{l}^{(1-2)}}.lite{{r}^{(-1+2)}}$
$K=mo{{l}^{-1}}.litre$
Thus the unit of equilibrium constant is $mo{{l}^{-1}}.litre$.
Thus the correct option is B.
Note: Units of equilibrium constant K will depend on the equilibrium of the concentration of the number of reactants to the products. If the total number of moles of product is equal to the total number of moles of reactant then the equilibrium constant K has no units i.e. it is dimensionless.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

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

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

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

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

NEET 2025 – Every New Update You Need to Know

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
