Question

At a certain temperature, this reaction establishes an equilibrium with the given equilibrium constant, Kc. 3 A ( g ) + 2 B ( g ) − ⇀ ↽ − 4 C ( g ) K c = 2.13 × 10 17 If, at this temperature, 1.80 mol of A and 3.90 mol of B are placed in a 1.00 L container, what are the concentrations of A, B, and C at equilibrium?

Answer 1

Answer:

The concentrations of A, B, and C at equilibrium:

[A] = 0.0 M

[B] =  2.7 M

[C] = 2.4 M

Explanation:

Concentration of 1.80 mol of A in 1.00 L container :

$[A]=\frac{1.80 mol}{1.00 L}=1.80 M$

Concentration of 3.90  mol of B in 1.00 L container :

$[B]=\frac{3.90 mol}{1.00 L}=3.90 M$

$3A(g)+2B(g)\rightleftharpoons 4C(g), K_c=2.13\times 10^{17}$

Initially

1.80 M        3.90 M                     0

At equilibrium

(1.80-3x)M     (3.90-2x)                       4x

The expression of an equilibrium constant is given by :

$K_c=\frac{[C]^4}{[A]^3[B]^2}$

$2.13\times 10^{17}=\frac{(4x)^4}{(1.80-3x)^3\times (3.90-2x)^2}$

Solving for x:

x = 0.600

The concentrations of A, B, and C at equilibrium:

[A] = [1.80-3x]=[1.80-3 × 0.600]= 0 M

[B] = [3.90-2x] = [3.90-2 × 0.600] = 2.7 M

[C] = [4x] =[4 × 0.600 M] = 2.4 M