Question

A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and allows it to react at room temperature (20.0°C): SCl2 (g) + 2 C2H4 (g) ⇋ S(CH2CH2Cl)2 (g)At equilibrium, [S(CH2CH2Cl)2] = 0.35 M. Calculate Kp.

Answer 1

The value of
`K_p` is 0.02495.

Step-by-step explanation:

Initial concentration of
`SCL_2` gas = 0.675 M

Initial concentration of
`C_2H_4` gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

`SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)`

initially

0.675 M 0.973 M 0

At equilibrium ;

(0.675-0.35) M (0.973-2 × 0.35) M 0.35 M

The equilibrium constant is given as :

`K_c=([S(CH_2CH_2Cl)_2])/([SCl_2][C_2H_4]^2)`

`=(0.35 M)/((0.675-0.35) M* ((0.973-2 × 0.35) M)^2)`

`K_c=14.45`

The relation between
`K_p` and
`K_c` are :

`K_p=K_c* (RT)^(\Delta n)`

where,

`K_p` = equilibrium constant at constant pressure = ?

`K_c` = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

`\Delta n` = change in the number of moles of gas = [(1) - (1 + 2)]=-2

Now put all the given values in the above relation, we get:

`K_p=14.45* (0.0821L.atm/K.mol* 293.15 K)^(-2)`

`K_p=6.2* 10^(4)`

`K_p=0.02495`

The value of
`K_p` is 0.02495.