Question

For the reaction given below, 2.00 moles of A and 3.00 moles of B are placed in a 6.00-L container. A(g) + 2B(g) C(g) At equilibrium, the concentration of A is 0.220 mol/L. What is the value of K? (6.90)

Answer 1

[A] = 2 mol / 6 L = 0.333 M

[B] = 3 mol/ 6 L = 0.5 M

A(g) + 2B(g) <-------------> C(g)

Initial 0.333 M 0.5 M 0

at equilibrium 0.333 -x 0.5-2x x

Hence,

equilibrium concentration of A = 0.333 -x

But given that equilibrium concentration of A = 0.220 M

Then, 0.333 -x = 0.22

x = 0.113 M

Therefore,

equilibrium concentrations are

[A] = 0.22 M

[B] = 0.5 - 2 x =0.5 - 2(0.113) = 0.274 M

[C] = x = 0.113 M

Then,

K = [C]/ [A] [B]²

= 0.113 / (0.22) (0.274)²

= 6.9

Therefore,

K = 6.90