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

Answer:

Kp = 0.81666

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,

$K_p=\frac{x^2}{0.500-x}$  

x = 0.350 atm

Thus,

$K_p=\frac{{0.350}^2}{0.500-0.350}$  

Thus, Kp = 0.81666