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.

The equilibrium concentration of methane, carbon tetrachloride and are 0.2686 M, 0.2686 M and 0.0828 M respectively.

Explanation:

Molarity is calculated by using the equation:

Moles of methane = 0.310 moles

Volume of solution = 1.00 L

Moles of carbon tetrachloride = 0.310 moles

Volume of solution = 1.00 L

For the given chemical equation:

Initial:          0.310        0.310

At eqllm:   0.310-x     0.310-x            2x

The expression of for above equation follows:

We are given:

Putting values in above equation, we get:

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 = 2x = (2 × 0.0414) = 0.0828 M

Hence, the equilibrium concentration of methane, carbon tetrachloride and are 0.2686 M, 0.2686 M and 0.0828 M respectively.

- both b and d.

-since b and d are the closest coastlines then that makes it the most affected.

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