Question

The equilibrum constant K_c is 0.01323 for the reaction: CCl4 _(g) <----> C_(s) + 2 Cl_{2 (g)}

At 300k a 5L flask originally contained 0.0828M of CCl_4, 0.0444 M of C and 0.0546M of Cl_2. Determine the concentration of Cl₂ when equilibrium is reached.

Answer 1

we only consider liquid and gas concentration in calculation

CCl_4(g) ---> C_(s) + 2Cl_2(g)

Initial 0.0828 0.0546

Final (0.0828-x) (0.0546+2x)

K_c = [Cl₂]²/[CCl₄]

0.01323 = (0.0546+2x)²/(0.0828-x)

4x²+0.2184x+0.003 = 0.0011-0.01323x

4x²+0.23163x+0.0019=0

x = -0.0099 = -0.01 M which is acceptable or x = -0.0480 M which is not accpetable because from this x, [Cl₂] will become negative.

so, x = -0.01 M

so reaction will go in backward direction.

Cl₂ concentration = 0.0546 +2(-0.01) = 0.0346 M