For the exothermic reaction
PCl3(g)+Cl2(g)?PCl5(g)
Kp = 0.180 at a certain temperature.
A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300atm PCl5 at this temperature.
What are the equilibrium partial pressures of PCl3 , Cl2, and PCl5, respectively?

Explanation:

1. Set up an ICE table

2. Write the Kp expression

3. Substitute the partial pressures and solve for x

We reject the positive answer, because, even if the reaction goes completely to the right, the partial pressure of PCl₅ cannot exceed 0.8 atm.

The negative sign shows that the reaction goes in the reverse direction.

Check:

Close enough.

pPCl3 =  0.709 atm

pCl2 = 0.709 atm

pPCl5 = 0.091 am

Explanation:

Step 1: Data given

Kp = 0.180

A flask is charged with 0.500 atm PCl3 , 0.500 atm Cl2, and 0.300atm PCl5 at this temperature

Step 2: The balanced equation

PCl3(g)+Cl2(g) ⇆ PCl5(g)

Step 3: The pressure at the equilibrium

pPCl3 = 0.500 - X atm

pCl2 = 0.500 - X atm

pPCl5 = 0.300 + X am

Step 4: Calculate the partial pressures

Kp = (pPCl5) / (pPCl3) *(pCl2)

0.180 = (0.300+X) / (0.500-X) (0.500-X)

0.180 = (0.300+X) / (0.250 - X + X^2)

X = -0.209

pPCl3 = 0.500 +0.209 = 0.709 atm

pCl2 = 0.500 + 0.209 = 0.709 atm

pPCl5 = 0.300 - 0.209 = 0.091 am

Kp = 0.091 / (0.709*0.709)

Kp = 0.18

i think so it’s will be b yes i believe its

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