2.848 mole of A (g) is placed in a 2.00 liter container at 600 °C. The equilibrium constant is 0.441. Find the concentrations of the reactant and both products once equilibrium has been reached.

Question

The equation is 2 A (g) ⇋ B (g) + C (g)

Answer 1

Solving this problem without using the quadratic equation.

Starting with 2.848 moles of A(g) in a 2.0 L container Initially at 600 degrees centigrade and Kc = 0.441

To find M: 2.848 mol/2.0 L = 1.424M

equation:

2A(g) ----> B(g) + C(g). using and ICE table (Initial, Change, Equilibrium)

<----

I 1.424 0 0

C -2x +x. +x

E 1.424-2x. x x

the equilibrium equation: Kc = [B][C]/[A]^2

A = 1.424-2x

B = x

C = x

0.441 = (x)(x)/(1.424-2x)^2 which is the equivalent to (x)^2/(1.424-2x)^2

Taking the square root of both sides to obtain x without an exponent =

0.664 = x/(1.424-2x) now solve for x

0.664(1.424-2x) = x. (distributing on the left side of the equation)

0.946-1.33x = x

0.946 = 2.33x

x = 0.406 M

plugging the value of x into our equilibrium section on our ICE table:

A(g) = 1.424 - (2)(0.406) = 0.612 M

B(g) = 0.406 M

C(g) = 0.406 M

Answer 2

Set up an ICE table

Initial [A] = 2.848 mol / 2.0 L = 1.424 M

2 A (g) ⇋ B (g) + C (g)

1.424.......0...........0..........Initial

-2x..........+x........+x..........Change

1.424-2x...x.........x...........Equilibrium

Write the equilibrium expression and fill in the valules:

Keq = 0.441 = [B][C] / [A]²

0.441 = (x)(x) / (1.424 - 2x)²

solving for x using the quadratic formula I obtain

x = 0.409 M

Final concentration at equilibrium would be...

[A] = 1.424 - 2x = 1.424 - 0.818 = 0.606 M

[B] = x = 0.409 M

[C] = x = 0.409 M

(be sure to check all the math)