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

Answer:

The value of $K_p$ is 0.02495.

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 :

$K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}$

$=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}$

$K_c=14.45$

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

$K_p=K_c\times (RT)^{\Delta n}$

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:

$K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}$

$K_p=6.2\times 10^{4}$

$K_p=0.02495$

The value of $K_p$ is 0.02495.