Question

At a certain temperature, the equilibrium constant, Kc,Kc, for this reaction is 53.3. H2(g)+I2(g)↽−−⇀2HI(g)Kc=53.3 H2(g)+I2(g)↽−−⇀2HI(g)Kc=53.3 At this temperature, 0.400 mol H20.400 mol H2 and 0.400 mol I20.400 mol I2 were placed in a 1.00 L container to react. What concentration of HIHI is present at equilibrium?

Answer 1

Answer:

0.628 mol·L⁻¹

Explanation:

The balanced equation is

H₂ + I₂ ⇌ 2I ₂

Data:

 Kc = 53.3

[H₂] = 0.400 mol·L⁻¹

[I₂] = 0.400 mol·L⁻¹

1. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{H}_{2}& + & \text{I}_{2} & \, \rightleftharpoons \, & \text{2HI} & & \\0.400 & & 0.400 & & 0 & & \\-x & & -x & & +2x & & \\0.400 - x & & 0.400 - x & & 2x & & \\\end{array}$

2. Calculate the equilibrium concentrations

$K_{\text{c}} = \dfrac{\text{[HI] ^{2} }}{\text{[H _{2} ][I _2 ]}} = \dfrac{(2x)^{2}}{(0.400 - x)^{2}} = 53.3\\\\\begin{array}{rcl}\dfrac{(2x)^{2}}{(0.400 - x)^{2}} &=& 53.3\\ \dfrac{2x }{0.400 - x} & = & 7.301\\\\2x & = & 7.301(0.400 - x)\\2x & = & 2.920 - 7.301x\\9.301x & = & 2.920\\x & = & \dfrac{2.920}{9.301}\\\\x & = & \mathbf{0.3140}\\\end{array}$

[HI] = 2x mol·L⁻¹ = 2 × 0.3140 mol·L⁻¹ = 0.628 mol·L⁻¹