In an experiment the equilibrium constant for the reaction is K when the initiatief concentration of A and B is 0.1. Under the similar conditions in an another experiment if the initial concentration of A and B is 2 and 3 then find the value of equilibrium constant?

Question

Answer

Calculating Equilibrium Constant

Equilibrium constant (K) is defined as the ratio of the concentration of reactants and products at equilibrium. It is a constant value for a particular reaction at a specific temperature. The equilibrium constant can be calculated using the following formula:

K = [C]^c[D]^d/[A]^a[B]^b

where [A], [B], [C], and [D] are the concentrations of the reactants and products at equilibrium and a, b, c, and d are the coefficients of the balanced chemical equation.

Solution

In the first experiment, the initial concentration of A and B is 0.1. Let us assume that the balanced chemical equation for the reaction is:

aA + bB ↔ cC + dD

where a, b, c, and d are the coefficients of the balanced chemical equation.

Let us assume that at equilibrium, the concentration of A, B, C, and D is [A], [B], [C], and [D], respectively.

Using the formula for equilibrium constant, we get:

K = [C]^c[D]^d/[A]^a[B]^b = [C]^c[D]^d/(0.1)^a(0.1)^b

Similarly, in the second experiment, the initial concentration of A and B is 2 and 3, respectively.

Using the formula for equilibrium constant, we get:

K' = [C']^c[D']^d/[A']^a[B']^b = [C']^c[D']^d/(2)^a(3)^b

We can see that the only difference between K and K' is the concentration of the reactants and products at equilibrium. To find the value of K' in terms of K, we can use the concept of stoichiometry.

Stoichiometry tells us that the ratio of the moles of reactants and products in a balanced chemical equation is fixed. Therefore, if we know the initial concentrations of the reactants and products, we can use stoichiometry to find the concentrations of the reactants and products at equilibrium.

Let us assume that the moles of A and B in the first experiment are nA and nB, respectively. Similarly, let us assume that the moles of A and B in the second experiment are nA' and nB', respectively.

Using stoichiometry, we can write:

nA/nB = nA'/nB'

Therefore,

nA'/nA = nB'/nB = 2/0.1 = 20

This means that the concentration of A and B in the second experiment is 20 times higher than the concentration of A and B in the first experiment.

Using the formula for equilibrium constant, we get:

K' = [C']^c[D']^

Answer

Equilibrium constant K depends only on temperature change so change in concentration of reactant doesnot effect the value of K

SO EQUILIBRIUM CONSTANT value will remain unchanged

Similar NEET Doubts

In a reaction A + B C + D, the initial concentrations of A and B were 0.9 mol dm–3 each. At equilibrium the concentration of D was found to be 0.6 mol dm–3. What is the value of equilibrium constant for the reaction

What is the concentration of O2 in a fresh water stream in equilibrium with air at 25ºC and 1.0 bar ? Given, KH (Henry's law constant) of O2 = 1.3 × 10–3 mol/kg bar at 25ºC. (

1 mole of x and Y are allowed to react in a 2 litre container when equilibrium is reached the following reaction occur 2x Y=Z. If the concentration of Z is .2M calculate the equilibrium constant for this reaction?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

Physics Class 11

Top Courses for NEET

Top Courses for NEET

Suggested Free Tests

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company