Question

What is the %-yield of AB at equilibrium for the following reaction if 1.00 mol of each reactant was placed in a 10.0 L flask and heated to 500°C ? A2(g) + B2(g) ↔ 2AB(g) Kc = 0.25 at 500°C

Answer 1

V = 10 L then

if

K = [AB]^2 / [A2][B2]

[A2] = 1/10 = 0.1

[B2] = 1/10 = 0.1

[AB] = 0

in equilibrium

[A2] = 1/10 = 0.1 - x

[B2] = 1/10 = 0.1 - x

[AB] = 0 + 2x

K = [AB]^2 / [A2][B2]

0.25 = (2x)^2 / (0.1-x)^2

solve for x

sqrt(0.25) = 2x/(0.1-x)

0.5 = 2x/(0.1-x)

0.05 - 0.5x = 2x

2.5x = 0.05

x = 0.05/2.5 = 0.02

then

[A2] = 1/10 = 0.1 - 0.02= 0.08

[B2] = 1/10 = 0.1 - 0.02= 0.08

[AB] = 0 + 2x= 2*0.08 = 0.16

then

theoretical yield woulb be... 2*0.1 = 0.20

actual yield = 0.16

% yield = real/theoretical ¨100 = 0.16/0.20 *100 = 80%