Respuesta :
Answer:
The value of [tex]K_p[/tex] is 0.02495.
Explanation:
Initial concentration of [tex]SCL_2[/tex] gas = 0.675 M
Initial concentration of [tex]C_2H_4[/tex] gas = 0.973 M
Equilibrium concentration of mustard gas = 0.35 M
[tex]SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)[/tex]
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 :
[tex]K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}[/tex]
[tex]=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}[/tex]
[tex]K_c=14.45[/tex]
The relation between [tex]K_p[/tex] and [tex]K_c[/tex] are :
[tex]K_p=K_c\times (RT)^{\Delta n}[/tex]
where,
[tex]K_p[/tex] = equilibrium constant at constant pressure = ?
[tex]K_c[/tex] = 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
[tex]\Delta n[/tex] = change in the number of moles of gas = [(1) - (1 + 2)]=-2
Now put all the given values in the above relation, we get:
[tex]K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}[/tex]
[tex]K_p=6.2\times 10^{4}[/tex]
[tex]K_p=0.02495[/tex]
The value of [tex]K_p[/tex] is 0.02495.
