Question

Carbon disulfide and chlorine react according to the following equation: CS₂(g) + 3Cl2(g) S₂Cl₂(g) + CCl4(g) When 2.14 mol of CS₂ and 5.85 mol of Cl₂ are placed in a 2.00-L container and allowed to come to equilibrium, the mixture is found to contain 0.620 mol of CCl4. How many moles of Cl₂ are present at equilibrium? A) 1.520 mol B) 0.620 mol C) 3.99 mol D) 4.61 mol​

Answer 1

A) 1.520 mol

Step-by-step explanation:

n = PV/RT = (Ptotal - Pother) V/RT

PV = nRT

n = PV/RT = (Ptotal - Pother) V/RT

n(Cl₂) = [(5.85 - 3x) / 2.00] (0.08206) (298) / (1.00)

n(Cl₂) = 0.03682 (5.85 - 3x)

Kc = [S₂Cl₂] [CCl₄] / [CS₂] [Cl₂]^3

Kc = (0.620/x)(x/3) / [(2.14 - x)(5.85 - 3x)^3]

Kc = 0.009376 / (2.14 - x)(5.85 - 3x)^3

Kc = 1.78 = 0.009376 / (2.14 - x)(5.85 - 3x)^3

x = 1.52 mol