For the reaction below, Kc = 4.60 × 10⁻⁶. Note Kc is sometimes called K. What is the equilibrium concentration of B if the reaction begins with 0.550 M A? A (g) ⇌ 3 B (g)

Answer 1

A(g) <==> 3B(g)

0.550.........0...........Initial

-x..............+3x.......Change

0.550-x......3x.........Equilibrium

Kc = [B]³ / [A]

4.6x10^-6 = (3x)³ / 0.550 - x (assume x is small relative to 0.550 and ignore it in denominator)

4.6x10^-6 = (3x)³ / 0.550

27x³ = 2.53x10^-6

x³ = 9x10^-8

x = 4.54x10^-3 M (note this is less than 5% of 0.550 so above assumption was valid)

Equilibrium [B] = 3x

Equilibrium [B] = (3)(4.54x10^-3 M)

Equilibrium [B] = 0.0136 M