Question

A sample of PCl5(g) was placed in an otherwise empty flask at an initial pressure of 0.500 atm and a temperature above 500 K. Over time the PCl5 decomposed to PCl3(g) and Cl2(g): PCl5(g) PCl3(g) + Cl2(g) At equilibrium the pressure of PCl5 in the flask was found to be 0.150 atm. Calculate the value of Kp for this reaction at this temperature. Kp =

Answer 1

Kp = 0.81666

Step-by-step explanation:

Pressure of PCl₅ = 0.500 atm

Considering the ICE table for the equilibrium as:

PCl₅ (g) ⇔ PCl₃ (g) + Cl₂ (g)

t = o 0.500

t = eq -x x x

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Moles at eq: 0.500-x x x

Given the pressure of PCl₅ at equilibrium = 0.150 atm

Thus, 0.500 - x = 0.150

x = 0.350 atm

The expression for the equilibrium constant is:

`K_p=\frac {P_(PCl_3)P_([Cl_2)}{P_(PCl_5)}`

So,

`K_p=(x^2)/(0.500-x)`

x = 0.350 atm

Thus,

`K_p=\frac{{0.350}^2}{0.500-0.350}`

Thus, Kp = 0.81666