Question

The value of Kc for the reaction between water vapor and dichlorine monoxide, H2O(g) 1 Cl2O(g) 4 2 HOCl(g) is 0.0900 at 25°C. Determine the equilibrium concentrations of all three compounds at 25°C if the starting concentrations of both reactants are 0.00432 M and no HOCl is present.

Answer 1

Answer:

[HOCl] = 0.001 127 mol·L⁻¹; [H₂O] = [Cl₂O] = 0.003 76 mol·L⁻¹

Explanation:

The balanced equation is

H₂O + Cl₂O ⇌ 2HOCl

Data:

     Kc = 0.0900

[H₂O] = 0.004 32 mol·L⁻¹

[Cl₂O] = 0.004 32 mol

1. Set up an ICE table.

$\begin{array}{ccccccc}\rm \text{H _{2} O}& + & \text{Cl _{2} O} & \, \rightleftharpoons \, & \text{2HOCl} & & \\0.00432 & & 0.00432 & & 0 & & \\-x &&-x&&+2x&&\\0.00432-x &&0.00432 - x& & 2x&&\\\end{array}$

2. Calculate the equilibrium concentrations

$K_{\text{c}} = \dfrac{\text{[HOCl] ^{2} }}{\text{[H _{2} O][Cl _2 O]}} = \dfrac{(2x)^{2}}{(0.00432 - x)^{2}} = 0.0900\\\\\begin{array}{rcl}\dfrac{4x^{2}}{(0.00432 - x)^{2}} &=& 0.0900\\ \dfrac{2x }{0.00432 - x} & = & 0.300\\2x & = & 0.300(0.00432 - x)\\2x & = & 0.001296 - 0.300x\\2.300x & = & 0.001296\\x & = & \mathbf{5.63\times 10^{-4}}\\\end{array}$

[HOCl] = 2x mol·L⁻¹ = 2 × 5.63 × 10⁻⁴ mol·L⁻¹ =0.001 127 mol·L⁻¹

[H₂O] = [Cl₂O] = (0.004 32 - 0.000 563) mol·L⁻¹ = 0.003 76 mol·L⁻¹

Check:

$\begin{array}{rcl}\dfrac{0.001127^{2}}{0.00376^{2}} & = & 0.0900\\\\\dfrac{1.270 \times 10^{-6}}{1.411 \times 10^{-5}} & = & 0.0900\\0.0900 & = & 0.0900\\\end{array}$

OK.

Answer 2

Kc gives the equilibrium constant of the moles of the substance. The concentration of HOCl = 0.001127 mol per L, and of water and dichlorine monoxide = 0.00376 mol per L.

What is equilibrium concentrations?

Equilibrium concentrations are the total of the initial concentration and the change in the concentrations. It is given by the ICE table. See the attached image below for the ICE table.

The equilibrium concentrations from the ICE table can be calculated as:

$\begin{aligned}\rm Kc &= \rm \dfrac{[HOCl]^{2}}{[H_{2}O][Cl_{2}O]} \\\\\rm Kc &= \rm \dfrac {(2x^{2})}{(0.00432-x^{2})} = 0.0900\end{aligned}$

Solving further:

$\begin{aligned} 0.300 &= \rm \dfrac{2x}{0.00432-x}\\\\\rm 2x &= \rm 0.001296 - 0.300x\\\\\rm x &= 5.63\times 10^{-4}\end{aligned}$

From the ICE table, the individual concentration is calculated as:

$\begin{aligned} \rm [HOCl] &= \rm 2x \\\\&= 2 \times 5.63 \times 10^{-4}\\\\&=0.001127 \;\rm mol L^{-1}\end{aligned}$

And,

$\begin{aligned} \rm [H_{2}O] = [Cl_{2}O] &= (0.00432 - 0.000563) \\\\&= 0.00376 \;\rm mol L^{-1}\end{aligned}$

Therefore, the equilibrium constant can be used for the determination of the concentrations.

Learn more about equilibrium constants here:

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