Question

1.00 mol of A and 1.00 mol of B are placed in a 4.00-L container. After equilibrium is established, 0.400 mol of D is present in the container. Calculate the equilibrium constant for the reaction: A(g) + 3 B(g) <---> C(g) + 2 D(g)

Answer 1

Solution

Given data

volume of container = 4.00 L

initial moles of A = 1.00 mol

initial moles of B= 1.00 mol

moles of D at equilibrium = 0.400 mol

balanced reaction equation is as follows

A(g) +3B(g)   <----> C(g) + 2D(g)

Kc= ?

lets first calculate the concnetration of the A, B and D

initial concetration of A = 1.00 mol / 4.00 L = 0.25 M

Initial concnetration of B = 1.00 mol / 4.00 L = 0.25 M

Equilibrium Concentration of D = 0.400 mol / 4 L = 0.100 M

Now lets make the ICE table for the reaction

      A(g)    +    3B(g) <----->   C(g)   +    2D(g)

I   0.25 M       0.25 M              0               0

C -x               -3x                   +x            +2x

E 0.25-x       0.25-3x                x             0.100 M

now as we know the equilibrium concentration of the D that is 2x = 0.100 M

therefore x= 0.100 / 2 = 0.05 M

now using this value of x calculate the equilibrrium concentrations of the A ,B and C

[A] equi = 0.25 M - x = (0.25 M-0.05 M) = 0.20 M

[B] equi = 0.25 M - 3x = 0.25 M - (3*0.05) = 0.100 M

[C] equi = x = 0.05 M

now using the equlibrium concentrations of the each species calculate the equlibrium constant

Kc= [C][D]²/[A][B]³

Kc = [0.05] [0.100]²/[0.20][0.100]³

kc = 2.5

Equlibrium constant for the reaction = 2.5