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
[tex]K_{\text{eq}} = \dfrac{\text{[HI]$^{2}$}}{\text{[H$_{2}]$[I$_{2}$]}} = \dfrac{2.83^{2}}{0.335 \times 0.335} = 65.4[/tex]
2. Set up an ICE table.
[tex]\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}[/tex]
Step 3. Calculate the equilibrium concentrations
[tex]K_{\text{c}} = \dfrac{\text{[HI]$^{2}$}}{\text{[H$_{2}$][I$_2$]}} = \dfrac{(3.83 - 2x)^{2}}{(0.335 + x)^{2}} = 65.4[/tex]
[tex]\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}[/tex]
[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:
[tex]\begin{array}{rcl}\dfrac{3.61^{2}}{0.446^{2}} & = & 65.4\\\\65.5 & = & 65.4\\\end{array}[/tex]
Close enough.
