Suppose 75.0 g of PCl₅(g)is introduced into a 3.00 L vessel containing 10.0 g of Cl2(g),and the system is allowed to reach equilibrium at 250ºC. PCl₅(g) PCl₃(g)+ Cl₂(g) If K_P = 1.80 for this reaction,what is the total pressure inside the vessel at equilibrium? (R = 0.08206 L • atm/K • mol)

Question

Quizplus · Accepted Answer

First, we need to set up the ICE table to determine the equilibrium concentrations:

|       | PCl5    | Cl2     |      |       |
|-------|---------|---------|------|-------|
|Initial| 1.00 M  |  0.00 M |      |       |
|Change | -x      |  -x     |  +x  |  +x   |
|Equil. | 1.00-x  |  0.00-x |  x   |  x    |

Substitute the concentrations into the equilibrium expression and solve for x:

K = [PCl5]/[Cl2]
1.80 = (x)(x)/(1.00-x)
x = 0.990 M

Now, we can calculate the partial pressures of each gas at equilibrium (assuming ideal gas behavior):

PPCl5 = (1.00 - 0.990) * 0.08206 * 523.15 / 3.00 = 0.568 atm
PCl3 = x * 0.08206 * 523.15 / 3.00 = 4.79 atm
PCl2 = (0.0100 + 0.990 - x) * 0.08206 * 523.15 / 3.00 = 3.50 atm

Finally, we can calculate the total pressure inside the vessel at equilibrium:

Ptotal = PCl5 + PCl3 + PCl2 = 0.568 + 4.79 + 3.50 = 8.86 atm

Therefore, the answer is B.