Answer:
0.0097 mol·L⁻¹
Explanation:
The balanced equation is
2NOCl ⇌ 2NO₂ + Cl₂
Data:
Your value of Kc is incorrect. It should be
Kc =1.6 × 10⁻⁵
[NOCl] = 0.50 mol·L⁻¹
[NO] = 0.00 mol·L⁻¹
[Cl₂] = 0.00 mol·L⁻¹
Calculations:
1. Set up an ICE table.
[tex]\begin{array}{ccccccc}\rm \text{2NOCl}& \, \rightleftharpoons \, & \text{2NO} & +&\text{Cl}_{2} \\0.50 & &0.00 & & 0.00 & & \\-2x & & +2x & & +x & & \\0.50 -2x & & 0.00 + 2x & & 0.00 + x & & \\\end{array}[/tex]
2. Calculate the equilibrium concentrations
[tex]K_{\text{c}} = \dfrac{\text{[NO]$^{2}$[Cl$_{2}$]}}{\text{[NOCl]}^{2}} = \dfrac{(2x)^{2}(x)}{(0.50 - 2x)^{2}} = 1.6 \times 10^{-5}\\\\4x^{3} = 1.6 \times 10^{-5}(0.50 - 2x)^{2}\\x^{3} = 4.0 \times 10^{-6}(0.50 - 2x)^{2}[/tex]
This is a cubic equation. Some calculators can solve cubic equations, but we can solve it by the method of successive approximations.
We will make changes to the right-hand side until the calculated value of x no longer changes,
(a) 1st approximation
Assume that 2x is negligible compared to 0.50. Then
[tex]x = \sqrt [3] {4.0 \times 10^{-6}(0.50)^{2}} = 0.010[/tex]
(b) 2nd approximation
Assume that x= 0.010. Then
[tex]x = \sqrt [3] {4.0 \times 10^{-6}(0.50 - 2\times 0.010)^{2}} = 0.0097[/tex]
(b) 3rd approximation
Assume that x= 0.0097. Then
[tex]x = \sqrt [3] {4.0 \times 10^{-6}(0.50 - 2\times 0.0097)^{2}} = 0.0097[/tex]
No change, so x = 0.0097.
[Cl₂] = x mol·L⁻¹ = 0.0097 mol·L⁻¹
Check:
[tex]\begin{array}{rcl}\dfrac{(2\times 0.0097)^{2} \times 0.0097}{(0.050 - 2 \times0.097)^{2}} & = & 1.6 \times 10^{-5}\\\\\dfrac{3.76 \times 10^{-4} \times 0.0097}{(0.050 - 0.0194)^{2}} & = & 1.6 \times 10^{-5}\\\\\dfrac{3.65 \times 10^{-6}}{(0.481)^{2}} & = & 1.6 \times 10^{-5}\\\\\dfrac{3.65 \times 10^{-6}}{0.231} & = & 1.6 \times 10^{-5}\\\\1.6\times 10^{-5} & = & 1.6 \times 10^{-5}\\\end{array}[/tex]
It checks.
