Question

Consider the gas phase reversible reaction 2A ↔ B that occurs at atmospheric pressure.

a) If the equilibrium constant Ka = 0.1, what are the equilibrium mole fractions of A and B?

b) Assuming the reaction is elementary, write the rate law using kA and k-A as the forward and reverse rate constants, respectively. What is the equilibrium constant Kc in terms of kA and k-A? Assuming an ideal gas system, write an expression for Ka in terms of kA and k-A. If the temperature is 25oC, what is the value of KC?

c) Assuming the reaction is elementary, and that the forward reaction rate constant kA is 2.4 L/(mol·min), determine the time for the reaction to reach 80% of equilibrium in a constant volume batch reactor. Use the value for KC you obtained in part d to solve the problem.

Answer 1

Solution:

(a) equalibrium constant Ka = a_B / a_A² Where a_B and a_A are activity of B and A, respectively.

a_B = For ideal gas and pressure = 1 atm = mole fraction * Partial pressure/ total pressure = mole fraction

0.1 = a_B / a_A²

if mole fraction of X_A = x so X_B = 1-x

For atmospheric pressure P=1;

So equation becomes: 1-x = 0.1 x²

0.1 x²+x-1 = 0

Solving above quardratic equation will give:

Mole fraction of A = 0.92

Mole fraction of B = 1-0.92 = 0.08

(b) Rate of forward reaction (r_forw ) = K_A [A]²

Rate of backward reaction (r_back ) = K_A-A[B]

At equalibrium:

Rate of forward = Rate of backward

K_A [A]² = K_A-A[B]

equalibrium constant (Kc) = K_A/K_A-A = [B]/[A]²

thermodynamic equalibrium (Ka) = a_B / a_A² = (P_B/P_B⁰)/(P_A / P_A⁰)² =K_A/K_A-A

We know that: P = CRT; standard state pressure = P_A⁰ = P_B⁰ = 1 atm

So, (P_B/P_B⁰)/(P_A / P_A⁰)² = [B]RT/[A]²(RT)² = ([B]/[A]² ) RT-¹

^So Ka = ((Kc) or K_A/K_A-A)RT-¹

Kc = Ka (RT) = 0.1* 0.082 * 298 = 2.4436

Kc = 2.4436

(C) Given:

KA = 2.4 L/mol.min

Kc = 2.4436 (calculated in above part)

Conversion (x) = 0.8

Constant volume batch reactor:

Kc = K_A / K_A-A

So K_A-A = K_A/K_c = 0.982

C_B⁰ = 0 and let C_A⁰ = P/RT = 1/(0.082*298) = 0.041 mol/l

                2A <-> B

initial             0.041    0

After t time 0.041-x      x/2

t = 0.388 min. or 23.28 sec