Question

Kc = 0.040 for the system below at 450 oC: PCl5(g) ⇌ PCl3(g) + Cl2(g) Evaluate
Kp for the reaction at 450 oC.
(a) 0.40
(b) 0.64
(c) 2.4
(d) 0.052
(e) 6.7 x 10-4

Answer 1

Given: Kc = 0.040 for the system below at 450°C for the reaction $PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)$.

To find: We have to find the value of Kp.

Solution:

The given reaction is $PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)$.

We know that-

$Kp=Kc×(RT)^{∆n}$

Here Kc is 0.040.

R is the Rydberg constant which is equal to 0.082L.atm/(K.mol).

T is the temperature of the reaction which is equal to 450+273=723K.

∆n is the change in the number of moles given as-

∆n=2-1

∆n=1

Putting these values in the above formula we get-

$Kp = 0.040 \times {(0.082 \times 723)}^{1} \\ Kp = 0.040 \times 59.286 \\ Kp = 2.37$

The value of Kp is 2.37.

The correct option is c.