Observe the following equilibrium in a 1 L flask. At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.

Initial equilibrium: A leftharpoons B [A] = 0.5 M, [B] = 0.375 M Now 0.1 mole of A is added to the 1 L flask. New initial [A] = 0.5 + 0.1 = 0.6 M. Reaction shifts forward: A leftharpoons B Eq: 0.6 - x 0.375 + x 1.75x = 0.075 New concentration of A = 0.6 - 0.043 = 0.557 M New concentration of B = 0.375 + 0.043 = 0.418 M Le Chatelier addition problems require maintaining Keq constant. Set up ICE table after perturbation, equate to initial Keq. Class 11 Chemistry: Equilibrium

Observe the following equilibrium in a 1 L flask. At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to es

Initial equilibrium: A leftharpoons B [A] = 0.5 M, [B] = 0.375 M Now 0.1 mole of A is added to the 1 L flask. New initial [A] = 0.5 + 0.1 = 0.6 M. Reaction shifts forward: A leftharpoons B Eq: 0.6 - x 0.375 + x 1.75x = 0.075 New concentration of A = 0.6 - 0.043 = 0.557 M New concentration of B = 0.375 + 0.043 = 0.418 M Le Chatelier addition problems require maintaining Keq constant. Set up ICE table after perturbation, equate to initial Keq. Class 11 Chemistry: Equilibrium

Observe the following equilibrium in a 1 L flask. At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.

Initial equilibrium: A leftharpoons B [A] = 0.5 M, [B] = 0.375 M Now 0.1 mole of A is added to the 1 L flask. New initial [A] = 0.5 + 0.1 = 0.6 M. Reaction shifts forward: A leftharpoons B Eq: 0.6 - x 0.375 + x 1.75x = 0.075 New concentration of A = 0.6 - 0.043 = 0.557 M New concentration of B = 0.375 + 0.043 = 0.418 M Le Chatelier addition problems require maintaining Keq constant. Set up ICE table after perturbation, equate to initial Keq. Class 11 Chemistry: Equilibrium

Observe the following equilibrium in a 1 L flask. At T(K), the equilibrium concentrations of A and B are 0.5 M and 0.375 M respectively. 0.1 moles of A is added into the flask and heated to T(K) to es

Initial equilibrium: A leftharpoons B [A] = 0.5 M, [B] = 0.375 M Now 0.1 mole of A is added to the 1 L flask. New initial [A] = 0.5 + 0.1 = 0.6 M. Reaction shifts forward: A leftharpoons B Eq: 0.6 - x 0.375 + x 1.75x = 0.075 New concentration of A = 0.6 - 0.043 = 0.557 M New concentration of B = 0.375 + 0.043 = 0.418 M Le Chatelier addition problems require maintaining Keq constant. Set up ICE table after perturbation, equate to initial Keq. Class 11 Chemistry: Equilibrium

JEE Main · Chemistry ↓ Falling

Equilibrium appeared 36 times across 3 years — 4.2% of Chemistry. This question is from Le Chatelier Principle And Equilibrium Calculation.

Year	2026	2025	2024	Total
Questions	10	18	8	36

Practice all Equilibrium questions → View full JEE Main weightage & trends →

Observe the following equilibrium in a 1 L flask. $A(g) leftharpoons B(g)$ $$A(g) \rightleftharpoons B(g)$$ At T(K), the equilibrium concentrations of A and B are $0.5 M$ $0.5\text{ M}$ and $0.375 M$ $0.375\text{ M}$ respectively. $0.1 moles$ $0.1\text{ moles}$ of A is added into the flask and heated to T(K) to establish the equilibrium again. The new equilibrium concentrations (in M) of A and B are respectively.

A. $(1) 0.367, 0.275$ $(1)\text{ 0.367, 0.275}$
B. $(2) 0.53, 0.4$ $(2)\text{ 0.53, 0.4}$
C. $(3) 0.742, 0.557$ $(3)\text{ 0.742, 0.557}$
D. $(4) 0.557, 0.418$ $(4)\text{ 0.557, 0.418}$

Solution & Explanation

Related Formula

Keq = [B]eq[A]eq

Core Logic

Initial equilibrium: $A leftharpoons B$ $A \rightleftharpoons B$ $[A] = 0.5 M$ $[A] = 0.5\text{ M}$, $[B] = 0.375 M$ $[B] = 0.375\text{ M}$

Keq = (0.375)/(0.5) = 0.75

Now $0.1 mole$ $0.1\text{ mole}$ of $A$ $A$ is added to the 1 L flask. New initial $[A] = 0.5 + 0.1 = 0.6 M$ $[A] = 0.5 + 0.1 = 0.6\text{ M}$. Reaction shifts forward: $A leftharpoons B$ $A \rightleftharpoons B$ Eq: $0.6 - x 0.375 + x$ $0.6 - x \quad 0.375 + x$

Step 1: Solve for x

Keq = 0.75 = (0.375 + x)/(0.6 - x)

0.75(0.6 - x) = 0.375 + x

0.45 - 0.75x = 0.375 + x

$1.75x = 0.075$ $1.75x = 0.075$

x = (0.075)/(1.75) = (3)/(70) \approx 0.043 M

Step 2: Calculate New Concentrations

New concentration of $A = 0.6 - 0.043 = 0.557 M$ $A = 0.6 - 0.043 = 0.557\text{ M}$ New concentration of $B = 0.375 + 0.043 = 0.418 M$ $B = 0.375 + 0.043 = 0.418\text{ M}$

Pattern Recognition

Le Chatelier addition problems require maintaining $Keq$ $K_{eq}$ constant. Set up ICE table after perturbation, equate to initial $Keq$ $K_{eq}$.

Chapter Mix

Class 11 Chemistry: Equilibrium

Reference Study Guides

Equilibrium

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve for x

Step 2: Calculate New Concentrations

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Chapter Mix

More Equilibrium Questions — jee_main_2026_28_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Solution & Explanation

Related Formula

Core Logic

Step 1: Solve for x

Step 2: Calculate New Concentrations

Pattern Recognition

Chapter Mix

Reference Study Guides

More Equilibrium Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Chapter Mix

More Equilibrium Questions — jee_main_2026_28_january_evening

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back