Answer:
The equilibrium constant in terms of concentration that is, [tex]K_c=3.6243\times 10^{3}[/tex] .
Explanation:
[tex]PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)[/tex]
The relation of [tex]K_c\& K_p[/tex] is given by:
[tex]K_p=K_c(RT)^{\Delta n_g}[/tex]
[tex]K_p[/tex]= Equilibrium constant in terms of partial pressure.=98.1
[tex]K_c[/tex]= Equilibrium constant in terms of concentration  =?
T = temperature at which the equilibrium reaction is taking place.
R = universal gas constant
[tex]\Delta n_g[/tex] = Difference between gaseous moles on product side and reactant side=[tex]n_{g,p}-n_{g.r}=1-2=-1[/tex]
[tex]98.1=K_c(RT)^{-1}[/tex]
[tex]98.1 =\frac{K_c}{RT}[/tex]
[tex]K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3} [/tex]
The equilibrium constant in terms of concentration that is, [tex]K_c=3.6243\times 10^{3}[/tex] .
The Kc for the reaction is obtained as 3620.
We have to apply the formula;
Kp =Kc(RT)^Δng
Where;
Kp = 98.1
Kc = ?
R= 0.082 LatmK-1mol-1
T = 450 K
Δng = 1 - 2 = -1
Now we have to substitute into the equation;
98.1  = Kc(0.082 × 450)^-1
98.1  = Kc/(0.082 × 450)
Kc = 98.1 (0.082 × 450)
Kc = 3620
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