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he equilibrium constant, Kc, for the following reaction is 9.52×10-2 at 350 K:

CH4(g) + CCl4(g)→ 2CH2Cl2(g)

Calculate the equilibrium concentrations of reactants and product when 0.310 moles of CH4 and 0.310 moles of CCl4 are introduced into a 1.00 L vessel at 350 K.

Respuesta :

Answer: The equilibrium concentration of methane, carbon tetrachloride and [tex]CH_2Cl_2[/tex] are 0.2686 M, 0.2686 M and 0.0828 M respectively.

Explanation:

Molarity is calculated by using the equation:

[tex]\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution}}[/tex]

  • For methane:

Moles of methane = 0.310 moles

Volume of solution = 1.00 L

[tex]\text{Molarity of methane}=\frac{0.310}{1}=0.310M[/tex]

  • For carbon tetrachloride:

Moles of carbon tetrachloride = 0.310 moles

Volume of solution = 1.00 L

[tex]\text{Molarity of }CCl_4=\frac{0.310}{1}=0.310M[/tex]

For the given chemical equation:

                  [tex]CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)[/tex]

Initial:          0.310        0.310

At eqllm:   0.310-x     0.310-x            2x

The expression of [tex]K_c[/tex] for above equation follows:

[tex]K_c=\frac{[CH_2Cl_2]^2}{[CH_4][CCl_4]}[/tex]

We are given:

[tex]K_c=9.52\times 10^{-2}[/tex]

Putting values in above equation, we get:

[tex]9.52\times 10^{-2}=\frac{(2x)^2}{(0.310-x)\times (0.310-x)}\\\\x=-0.0565,0.0414[/tex]

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of methane = (0.310 - x) = [0.310 - 0.0414] = 0.2686 M

Equilibrium concentration of carbon tetrachloride = (0.310 - x) = [0.310 - 0.0414] = 0.2686 M

Equilibrium concentration of [tex]CH_2Cl_2[/tex] = 2x = (2 × 0.0414) = 0.0828 M

Hence, the equilibrium concentration of methane, carbon tetrachloride and [tex]CH_2Cl_2[/tex] are 0.2686 M, 0.2686 M and 0.0828 M respectively.

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