consider the reaction

Question

A (aq) --> 2 B (aq) Kc = 7.85 x 10^-6 at 500 K

If a 2.40 M sample of A is heated to 500 K, what is the concentration of  at equilibrium?

Answer 1

consider the reaction

A (aq) --> 2 B (aq) Kc = 7.85 x 10^-6 at 500 K

If a 2.40 M sample of A is heated to 500 K, what is the concentration of  at equilibrium?

Hello Rebecca,

........A --> 2 B

I .....2.4..... 0

Δ ... -x .....+2x

Eq. 2.4-x...0+2x

K_eq = [B]²/[A]

7.8x10^-6 = [2x]²/[2.4-x]

7.8x10^-6[2.4-x] = [2x]²

7.8x10^-6[2.4-x] = [2x]²

1.9x10^-5 - 7.8x10^-6x = 4x²

0 = 4x² + 7.8x10^-6x - 1.9x10^-5

x = 0.00218

At equilibrium, [A] = 2.4 - x = 2.4 - 0.00218 = 2.397,

At equilibrium, [B] = 0 + 2x = 2(0.00218) = 0.00436

At equilibrium, the concentration of [B] = 4.4x10^-3 M, [A] = 2.397 M

Answer 2

A(aq) <==> 2B(aq) ... Kc = 7.85x10^-6

2.4 M...........0........Initial

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

2.4-x...........2x.......Equilibrium

Kc = [B]² / [A]

7.85x10^-6 = [2x]² / [2.4-x]

7.85x10^-6 = 4x² / 2.4-x

1.88x10^-5 - 7.85x10^-6x = 4x²

4x² + 7.85x10^-6x -1.88x10^-5 = 0

x = 0.0022 M

Concentration of B at equilibrium = 2x = 2x0.0022 = 0.0044 M