Question

In a reaction vessel @ 310 K, an equilibrium mixture is found to contain: 0.0370 M PCl₃, 0.0174 M Cl₂ and 1.316 M PCl₅.

PCl₃ (g) + Cl₂ (g)  ⇋ PCl₅ (g)

Calculate Kp for the reaction.

I did this, and used the equation (1.316 M PCl5) / (.0370 M Cl3 * .0174 M Cl2) = 2044 = 2040 with sig figs.

Then because the delta n is -1 because of the coefficients, I used the equation

(2040) / (.08206 * 310)^1. I got the answer of 80.

My professor said this was incorrect. The thing is, even when I use the equation he supposedly used to get his answer, I get my answer of ~80. Below is the work that he used to calculate Kp. Any help would be appreciated.

(My professor's answer):

Answer

126

Kc=1.316(0.0370)(0.0174)=2040

Kp=Kc(RT)Δn

Kp=2040(0.0821∗310)−1=126

Answer 1

Kc = 2040 both your answers are correct.

there is the difference in Kp value

Kp = Kc (RT) ^{$\Delta$ n}

Kp = (2040) (0.0821 x 310 )^-1

Kp = 80.15

Kp = 80

so professor answer is wrong . you are correct.