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.
Kc= 0.040 for the system below at 450oC:
PC15(g)<<==>>PC13(g)+ C12(g) evaluate kp for the reaction at 450oc
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