The reaction PCl3(g) + Cl2(g) PCl5(g) has Kp = 0.0870 at 300 ?C. A flask is char

Question

The reaction PCl3(g) + Cl2(g) PCl5(g) has Kp = 0.0870 at 300 ?C. A flask is charged with 0.5 atm PCl3, 0.5 atm Cl2, and 0.2 atm PCl5 at this temperature.

(a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium.

(b) Calculate the equilibrium partial pressures of the gases.

(c) Once in equilibrium, the volume of the system is doubled. What direction would the reaction proceed to reach equilibrium again? Calculate the new equilibrium partial pressures.

Explanation / Answer

Kp = [PCl5] / [PCl3] [Cl2]

Q = [0.20] / [0.50] [0.50] = 0.8
since the Q is too high as compareded to the Kp of 0.087, the reaction will shift to the left to decrease the numerator PCl5

..... ....... PCl3 (g) + Cl2 (g) <--> PCl5 (g)
was .......(0.50) ... (0.50) ... ...( 0.20)
will shift:.. (0.50+x) ... (0.50+x) ... ...( 0.20-x)

Kp = [PCl5] / [PCl3] [Cl2]

0.0870 = [0.20-x] / [0.50+x] [0.50+x]

0.0870 [(0.50+x)(0.50+x)] = [0.20-x]

0.0870 [0.25 + 1.0x + x2] = [0.20-x]

0.02175 + 0.087x + 0.087x2 = 0.2 -x

0.087X2 + 1.087x -0.17825 = 0

x= 0.162
------------------

PCl3 = 0.50 + 0.162

Your answer = 0.66 atm

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