For the reaction below, Kc = 1.10 × 10⁻⁴. Note Kc is sometimes called K. What is the equilibrium concentration of C if the reaction begins with 0.200 M A and 0.400 M B? 2 A (aq) + B (aq) ⇌ C (aq)

Answer 1

2 A (aq) + B (aq) ⇌ C (aq) ... use an ICE table

0.200..........0.400.......0.......Initial

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

0.2-2x........0.4-x...........x......Equilibrium

Kc = [C] / [A]²[B]

1.1x10^-4 = [C] / [0.2-x]²[0.4-x] and assume that x is small relative to 0.2 and 0.4 and ignore it.

1.1x10^-4 = [C] / (0.2)²(0.4)

[C] = 1.76x10^-6 M (note: this is small relative to 0.2 and 0.4 so our assumption above was correct. If it were not correct, we'd have to repeat the calculation using the quadratic formula to solve for [C].)

Answer 2

Start with the base equation for the equilibrium constant: K_c = [products] / [reactants]

K_c = [C] / ([A]²[B]), [A] is squared because the reaction has a coefficient of 2 for A.

Set up your ICE table to get the equilibrium concentrations of [A] = (0.2 - 2x), [B] = (0.4 - x), [C] = (x)

Since K_c is small, we can ignore the x in [A] and [B], giving K_c = [C] / ([A]²[B]), 1.1*10^-4 = x / (0.2² * 0.4)

Rearrange for x; x = 1.1*10^-4 * (0.2² * 0.4) = 1.76 * 10^-6

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