A2 (g) + B2 (g) ⇋ 2 AB (g) Kc is 116.4, initial concentrations are: [A2] = 0.200 M, [B2] = 0.200 M, [AB] = 0.770 M, what will be the equilibrium concentration of AB (in M)?

Question

Consider the following equilibrium:

A₂ (g) + B₂ (g) ⇋ 2 AB (g)

If the value of K_c for this reaction is 116.4, and the initial concentrations of each species are: [A₂] = 0.200 M, [B₂] = 0.200 M, [AB] = 0.770 M , what will be the equilibrium concentration of AB (in M)?

Answer 1

[AB] = 0.987 M at equilibrium

Step 1 — which way does it go? Compare Q with K.

All three species are already present, so you cannot assume the reaction runs forward. Compute the reaction quotient from the starting values:

Q = [AB]² / ([A₂][B₂]) = (0.770)² / [(0.200)(0.200)] = 0.593 / 0.0400 = 14.8

Q = 14.8 is less than K = 116.4, so there is not enough product yet — the reaction shifts forward, consuming A₂ and B₂ to make more AB.

This step decides the signs in your ICE table. Guess wrong and every change is backwards, which usually surfaces as a negative concentration at the end.

Step 2 — ICE table

         A₂      B₂      AB
I:  0.200  0.200  0.770
C:  −x     −x    +2x
E: 0.200−x  0.200−x  0.770+2x

Step 3 — use the perfect-square shortcut

K = (0.770 + 2x)² / (0.200 − x)² = 116.4

Both sides are squares, so take the square root of the whole equation instead of expanding a quadratic:

(0.770 + 2x) / (0.200 − x) = √116.4 = 10.79

0.770 + 2x = 2.158 − 10.79x  →  12.79x = 1.388  →  x = 0.109

Step 4 — answer what was asked

[AB] = 0.770 + 2(0.109) = 0.987 M

(And [A₂] = [B₂] = 0.200 − 0.109 = 0.091 M.)

Check by substituting back

K = (0.987)² / (0.091)² = 0.974 / 0.00837 = 116.4 ✓

Always do this on equilibrium problems — it catches a sign error, an arithmetic slip, or a wrong root in about fifteen seconds.

Two notes

• Do not use the small-x approximation here. K is large and x came out at 54% of the initial A₂ concentration — nowhere near negligible.
• The square-root shortcut works because A₂ and B₂ start at equal concentrations with equal coefficients, making the denominator a perfect square. Had they started unequal, you would have to expand and use the quadratic formula.