College

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas:

\[ \text{PCl}_3(g) + \text{Cl}_2(g) \leftrightharpoons \text{PCl}_5(g) \]

A 7.5-L gas vessel is charged with a mixture of \(\text{PCl}_3(g)\) and \(\text{Cl}_2(g)\), which is allowed to equilibrate at 450 K. At equilibrium, the partial pressures of the three gases are:

\[ P_{\text{PCl}_3} = 0.125 \, \text{atm} \]
\[ P_{\text{Cl}_2} = 0.155 \, \text{atm} \]
\[ P_{\text{PCl}_5} = 1.90 \, \text{atm} \]

Given \( K_p = 98.1 \), what is \( K_c \)?

Answer :

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

Learn more about equilibrium constant: https://brainly.com/question/12978582

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] .

Other Questions