Question

Mustard gas, used in chemical warfare in World War I, has been found to be an effective agent in the chemotherapy of Hodgkin's disease. It can be produced according to the following reaction: SCL2(g)+2C2H4(g)<-->S(CH2CH2Cl)2(g). An evacuated 5.0L flask at 20.0C is filled with 0.258 mol SCl2 and 0.592 mol C2H4. After equilibrium is established, 0.0349 mol of mustard gas is present. 1. What is the partial pressure of each gas at equilibrium? 2. What is K at 20C?

Answer 1

For the given reaction -

                 SCl₂(g)   +   2 C₂H₄ (g) <--> S(CH₂CH₂Cl)₂ (g).

I(M)       (0.258/5)       (0.592/5)                 0

C              -(0.0349/5)     - 2(0.0349/5)        +0.0349/5

Eq         0.04462            0.104                    0.00698

Thus -

At equilibrium -

[SCl2] = 0.04462 M , [C2H4] = 0.104, [S(CH₂CH₂Cl)₂ ] = 0.00698 M

If we convert it in to pressures by using P = CRT (P= nRT/V)

P(SCl2) = (0.04462)(0.0821)(293) = 1.0733 atm

P(C2H4) = (0.104)(0.0821)(293) = 2.50 atm

P(S(CH₂CH₂Cl)2) = (0.00698)(0.0821)(293) = 0.167 atm

Hence - Kp = P(S(CH₂CH₂Cl)₂) / (P(C₂H₄))² * P(SCl₂) = (0.167) / (1.073)(2.50)² = 0.0249