Kc= 0.040 for the system below at 450oC:

PC15(g)<<==>>PC13(g)+ C12(g) evaluate kp for the reaction at 450oc

1 answer

We use the relationship:

Kp = Kc(RT)^(∆n)

where ∆n is the change in moles of gas between reactants and products.

We are given Kc = 0.040 at 450°C, but we need to find Kp. We also need to determine the value of ∆n.

Looking at the balanced chemical equation:

PCl5(g) <<==>> PCl3(g) + Cl2(g)

There are 2 moles of gas on the left side (PCl5) and 2 moles of gas on the right side (PCl3 and Cl2). Therefore, ∆n = (2 - 2) = 0.

Now we can use the equation to solve for Kp:

Kp = Kc(RT)^(∆n) = Kc(RT)^0 = Kc

At 450°C, we have:

Kp = Kc = 0.040

Therefore, the value of Kp for the reaction at 450°C is 0.040.