Answer :
Answer:
[tex]0.024[/tex]
Explanation:
First we have to use the ICE table to find the equilibrium concentrations:
[tex]SCl_2_(_g_) + 2 C_2H_4_(_g_) <->S(CH_2CH_2Cl)_2_(_g_)[/tex]
I 0.675 M 0.973 M 0
C -X -2X +X
E 0.675-X 0.973-2X X
The equilibrium concentration is given by the problem:
X=0.35 M
Then we have to do the substractions:
[tex]SCl_2_(_g_)[/tex] = 0.325 M
[tex] C_2H_4_(_g_)[/tex] = 0.273 M
[tex]S(CH_2CH_2Cl)_2_(_g_)[/tex]= 0.35 M
With the equilibrium concentrations is posible to calculate Kc:
[tex]Kc=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}=\frac{[0.35]}{[0.325][0.273]^2}=14.45[/tex]
With the Kc value is posible calculate Kp using the equation:
Kp = Kc(RT)^(Δn)
Kp= Equilibrium Pressure constant
Kc= Equilibrium Concentration constant
R= 0.08206 L*atm/mol*K
T= Temperature
Δn=moles of products - moles of reactants
T= 25 ºC= 298.15 K
Δn = 1 mole of products [tex]S(CH_2CH_2Cl)_2_(_g_)[/tex]- 3 moles of reactants [tex]2 C_2H_4_(_g_)~and~SCl_2_(_g_)[/tex]= -2
[tex]Kp=~14.45~(0.082*298.15)^(^-^2^)= 0.024[/tex]
From the ICE table, the Kp of the system is 0.599.
The first step is to set up the ICE table of the reaction equation as follows;
SCl2(g) + 2 C2H4(g) ⇌ S(CH2CH2Cl)2(g)
I 0.675 0.973 0
C - x -2x +x
E 0.675 - x 0.973 - 2x 0.350
But x = 0.350 M
Hence; Kp = [0.350]/[0.675 - 0.350] [0.973 - 2(0.350)]
Kp = 0.350/0.325 × (0.273)^2
Kp = 14.4
Kp = Kc(RT)^Δng
Kp = 14.4(0.082 × 293)^-2
Kp = 14.4/(0.082 × 293)^2
Kp = 0.599
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