The equilibrium constant, Kc, for the following reaction is 56.0 at 278 K. 2CH2Cl2 (g) CH4 (g) + CCl4 (g) When a sufficiently large sample of CH2Cl2 (g) is introduced into an evacuated vessel at 278 K, the equilibrium concentration of CCl4 (g) is found to be 0.348 M. Calculate the concentration of CH2Cl2 in the equilibrium mixture.

21.5 M

Explanation:

1) Equilibrium equation (given):

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

2) Write the concentration changes when some concentration, A, of CH₂Cl₂ (g) sample is introduced into an evacuated (empty) vessel:

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

A - x x x

3) Replace x with the known (found) equilibrium concentraion of CCl₄ (g) of 0.348 M

2CH₂Cl₂ (g) ⇄ CH₄ (g) + CCl₄ (g)

A - 0.3485 0.348 0.348

4) Write the equilibrium constant equation, replace the known values and solve for the unknown (A):

Kc = [ CH₄ (g) ] [ CCl₄ (g) ] / [ CH₂Cl₂ (g) ]²

56.0 = 0.348² / A²

A² = 56.0 / 0.348² = 462.

A = 21.5 M ← answer