Question

At 500 °C, hydrogen iodide decomposes according to 2 HI ( g ) − ⇀ ↽ − H 2 ( g ) + I 2 ( g ) For HI ( g ) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [ H 2 ] = 0.335 M , [ I 2 ] = 0.335 M , and [ HI ] = 2.83 M . If an additional 1.00 mol of HI ( g ) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached?

Answer 1

Answer:

[HI] = 3.61 mol·L⁻¹; [H₂] = [I₂] = 0.446 mol·L⁻¹  

Explanation:

1. Calculate Keq

                     H₂   +      I₂    ⇌ 2HI

E/mol·L⁻¹:    0.335    0.335    2.83

$K_{\text{eq}} = \dfrac{\text{[HI] ^{2} }}{\text{[H _{2}] [I _{2} ]}} = \dfrac{2.83^{2}}{0.335 \times 0.335} = 65.4$

2. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{H}_{2}& + & \text{I}_{2} & \, \rightleftharpoons \, & \text{2HI} & & \\0.335 & & 0.335 & & 3.83 & & \\+x & & +x & & -2x & & \\0.335 + x & & 0.335 + x & & 3.83 - 2x & & \\\end{array}$

Step 3. Calculate the equilibrium concentrations

$K_{\text{c}} = \dfrac{\text{[HI] ^{2} }}{\text{[H _{2} ][I _2 ]}} = \dfrac{(3.83 - 2x)^{2}}{(0.335 + x)^{2}} = 65.4$

$\begin{array}{rcl}\dfrac{(3.83 - 2x)^{2}}{(0.335 + x)^{2}} &=& 65.4\\\\ \dfrac{3.83 - 2x}{0.335 + x} & = & 8.09\\\\3.83 - 2x & = & 8.09(0.335 + x)\\3.83 - 2x& = & 2.709 + 8.09x\\ 1.121 & = & + 10.09x\\x & = & \dfrac{1.121}{10.09}\\\\x & = & \math{0.1111}\\\end{array}$

[HI] = (3.83 - 2x) mol·L⁻¹ = (3.83 - 0.2223) mol·L⁻¹ = 3.61 mol·L⁻¹

[H₂] = [I₂] = (0.335 + x) mol·L⁻¹ = (0.335 + 0.1111) =  0.446 mol·L⁻¹  

Check:

$\begin{array}{rcl}\dfrac{3.61^{2}}{0.446^{2}} & = & 65.4\\\\65.5 & = & 65.4\\\end{array}$

Close enough.