Question

use the references to access important values if needed for this question.
the equilibrium constant, ( k_{c} ), for the following reaction is 10.5 at 350 k.
2 mathrm{ch}_{2} mathrm{cl}_{2}(g)
ightleftharpoons mathrm{ch}_{4}(g)+mathrm{ccl}_{4}(g)
if an equilibrium mixture of the three gases in a 18.9 l container at 350 k contains 0.440 mol of ( mathrm{ch}_{2} mathrm{cl}_{2}(g) ) and 0.339 mol of ( mathrm{ch}_{4} ), the equilibrium concentration of ( mathrm{ccl}_{4} ) is m.

Explanation:

Step1: Calculate the concentration of \(CH_2Cl_2\) and \(CH_4\)

Concentration \(c=\frac{n}{V}\).
For \(CH_2Cl_2\), \(n = 0.440\space mol\), \(V=18.9\space L\), so \(c_{CH_2Cl_2}=\frac{0.440}{18.9}\space M\approx0.0233\space M\).
For \(CH_4\), \(n = 0.339\space mol\), \(V = 18.9\space L\), so \(c_{CH_4}=\frac{0.339}{18.9}\space M\approx0.0179\space M\).

Step2: Use the equilibrium constant expression

The equilibrium constant expression for the reaction \(2CH_2Cl_2(g)
ightleftharpoons CH_4(g)+CCl_4(g)\) is \(K_c=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}\).
We know \(K_c = 10.5\), \( [CH_4]=0.0179\space M\), \( [CH_2Cl_2]=0.0233\space M\).
Let \( [CCl_4]=x\). Then \(10.5=\frac{0.0179\times x}{(0.0233)^2}\).

Step3: Solve for \(x\) (concentration of \(CCl_4\))

First, calculate \((0.0233)^2=0.00054289\).
Then the equation becomes \(10.5\times0.00054289 = 0.0179x\).
\(0.005690345=0.0179x\).
\(x=\frac{0.005690345}{0.0179}\space M\approx0.318\space M\).

Answer:

\(0.318\)