Question

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl3(g)+Cl2(g)⇌PCl5(g). A 7.5-L gas vessel is charged with a mixture of PCl3(g) and Cl2(g), which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are PPCl3 = 0.125atm , PCl2 = 0.155atm , and PPCl5 = 1.90atm Kp= 98.1 What is Kc?

Answer 1

Answer:

Kc is 3623

Explanation:

Write out the parameters given:

Volume = 7.5 L,   Kp = 98.1, Partial pressure of $PCl_{3}$ at equilibrium = 0.125 atm,  Partial pressure of $Cl_{2}$ at equilibrium = 0.155 atm, Partial pressure of $PCl_{5}$ at equilibrium = 1.90 atm,

R = 0.08206 L·atm/(mol·K), T = 450 K

Formula to calculate Kc is given below:

$K_{p} = K_{c}(RT)^{n}$

n is given by the difference between the moles of gaseous products (PCl5) & the moles of gaseous reactants (PCl3) (Cl2) as seen in the chemical reaction $PCl_{3}$ (g) + $Cl_{2}$ (g) ⇌ $PCl_{5}$ (g)

n = 1 - 2 = -1

Substitute the parameters into the equation to solve for Kc, we have:

98.1 = $K_c (0.08206 X 450)^{-1}$

Kc = 3622.5 ≅ 3623

Kc = 3623

Answer 2

Answer:

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .

Explanation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The relation of $K_c\& K_p$ is given by:

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

$K_p$= Equilibrium constant in terms of partial pressure.=98.1

$K_c$= Equilibrium constant in terms of concentration  =?

T = temperature at which the equilibrium reaction is taking place.

R = universal gas constant

$\Delta n_g$ = Difference between gaseous moles on product side and reactant side=$n_{g,p}-n_{g.r}=1-2=-1$

$98.1=K_c(RT)^{-1}$

$98.1 =\frac{K_c}{RT}$

$K_c=98.1\times 0.0821 L atm/mol K\times 450 K=3,624.30=3.6243\times 10^{3}$

The equilibrium constant in terms of concentration that is, $K_c=3.6243\times 10^{3}$ .