For a hypothetical chemical equilibrium A=2B C it was found that initially only A and B are present with mole fraction B equal to 1/3. At equilibrium Initial molefraction of A become 1/2 then Kc is?

Question

For a hypothetical chemical equilibrium A=>2B C it was found that initially only A and B are present with mole fraction B equal to 1/3. At equilibrium Initial molefraction of A become 1/2 then Kc is?

Answer

Calculation of Kc for Chemical Equilibrium A=>2B+C

Given Information

Chemical equilibrium: A => 2B + C

Initial mole fraction of B: 1/3

Final mole fraction of A: 1/2

Calculating Equilibrium Concentrations

Let us assume that at equilibrium, the moles of A, B and C are a, b, and c respectively. Then we can write the expression for Kc as:

Kc = ([B]^2[C])/[A]

At the start of the reaction, we have:

[A] = a/V (where V is the total volume of the system)

[B] = b/V

[C] = c/V

At equilibrium, we have:

[A] = (a - x)/V, where x is the amount of A that has reacted

[B] = (b + 2x)/V, where x is the amount of A that has reacted

[C] = (c + x)/V, where x is the amount of A that has reacted

Calculating the Value of Kc

Substituting the equilibrium concentrations into the expression for Kc, we get:

Kc = ((b+2x)^2(c+x))/((a-x)V)

Since the initial mole fraction of B is 1/3, we can write:

b/a = 1/3

Solving for b and substituting into the expression for Kc, we get:

Kc = ((1/3 + 2x/a)^2(c+x))/((1/2 - x/a)V)

Since the initial mole fraction of A becomes 1/2 at equilibrium, we can write:

x/a = (1/2 - 1/2)/(1 + 2 + 1) = 1/12

Substituting this value into the expression for Kc, we get:

Kc = ((1/3 + 2(1/12))^2(c+1/12))/((1/2 - 1/12)V) = 49c/V

Therefore, the value of Kc for the chemical equilibrium A => 2B + C is 49c/V.

Answer

Similar Class 11 Doubts

The equilibrium constant KC for the reaction A 2B gives rise to 3C is 2 into 10 to the power minus 3 what would be the equilibrium partial pressure of gas if initial pressure of gas if A and B are a and to ATM respectively?

Which of the following equilibria is not affected by change in volume of the flask [AIEEE-2002]

The equilibrium constant Kp for the reaction H2 +CO2¬H20 +CO is 4at 1660C .initially 0.80 mole of h2 & 0.80 mole of CO2 are injected into 5L flask .What is the equilibrium concentration of CO2? 1)0.533M. 2)0.0534M. 3)5.34M. 4)none of this.ans is 2 one .can any one explain how?

Q.A chemicle reaction is at equilibrium when.(

The value of K_{p} for the reaction, CO_{2}(g)+C(s)\rightleftharpoons2CO(g) is 3.0 at 1000 K. If initially \mathfrak{p}_{cO},=0.48 bar and r_{CO}=0 bar and pure graphite is present, calculate the equilibrium partial pressures of CO and CO_{2}.?

Top Courses for Class 11View all

Chemistry Class 11

CUET Mock Test Series

English Class 11

Chemistry Practice Tests: CUET Preparation

CUET Mock Test: Humanities Subjects 2025

Top Courses for Class 11

Top Courses for Class 11

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams