Question

The equilibrium constant Kc for the reaction
H2 (g) + Br2 (g) <=> 2 HBr (g)
is 2.18 ×× 10^6 at 730°C. Starting with 3.20 moles of HBr in a 12.0-L reaction vessel, calculate the concentrations of H2, Br2, and HBr at equilibrium.

Answer 1

$H_{2(g)} + Br_{2(g)} \ ^\leftarrow_\rightarrow \ 2HBr_{(g)}\\ k_c = 2.18 × 10^6 \\ T = 730°C$

number of moles of HBr = 3.20 mol

V = 12.0 L

molarity of HBr = 3.2/12 = 0.27M

Using Ice,

H + Br = 2HBr

I ---- 0M + 0M = 0.27M

C ----- +x +x = -2x

E ----- x + x = 0.27 -2x

$\begin{aligned} k_c &= \dfrac{[HBr]^2}{[H_2][Br_2]}\\ \\ 2.18 × 10^6 &= \dfrac{[0.27-x]^2}{[x][x]}\\ \\ 2.18 × 10^6 &= (\dfrac{0.27-x}{x})^2\\ \\ \sqrt{2.18 × 10^6} &= \dfrac{0.27-x}{x}\\ \\ 1476.5 &= \dfrac{0.27-x}{x}\\ \\ 1476.5x &= 0.27 - x\\ 1477.5x &= 0.27\\ x &= 0.000183M \end{aligned}$

$\begin{aligned} \therefore [H_2] &= x = 1.83 × 10^4M \\ [Br_2] &= x = 1.83 × 10^4M\\ [HBr] &= 0.27 - 2x = 0.27 - 2(0.000183) = 0.27M \end{aligned}$