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Mustard gas, used in chemical warfare in World War I, has been found to be an...

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Chemistry

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

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4.2 Ratings (889 Votes)

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
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