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Chapter 13: Problem 78

For the reaction $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ at \(600 . \mathrm{K}\), the equilibrium constant, \(K_{\mathrm{p}}\), is \(11.5 .\) Suppose that \(2.450 \mathrm{~g} \mathrm{PCl}_{5}\) is placed in an evacuated \(500 .-\mathrm{mL}\) bulb, which is then heated to \(600 . \mathrm{K}\). a. What would be the pressure of \(\mathrm{PCl}_{5}\) if it did not dissociate? b. What is the partial pressure of \(\mathrm{PCl}_{5}\) at equilibrium? c. What is the total pressure in the bulb at equilibrium? d. What is the percent dissociation of \(\mathrm{PCl}_{5}\) at equilibrium?

Short Answer

Expert verified
The short answers are: a. The pressure of PCl5 if it did not dissociate would be 11.562 atm. b. The partial pressure of PCl5 at equilibrium is 0.0188 atm. c. The total pressure in the bulb at equilibrium is 0.02828 atm. d. The percent dissociation of PCl5 at equilibrium is 20.13%.

Step by step solution

01

Calculate the initial moles of PCl5

We need to calculate the initial moles of PCl5 using its mass and molar mass. Molar mass of PCl5 = 1P + 5Cl = 1(30.97) + 5(35.45) = 208.22 g/mol Given mass = 2.45 g Initial moles of PCl5 = mass/molar mass = 2.45/208.22 = 0.01177 mol
02

Calculate the initial pressure of PCl5 if it did not dissociate

We can use the ideal gas law equation to find the pressure of PCl5 assuming it doesn't dissociate, PV = nRT Where P is the pressure, V is the volume (0.5 L), n is the moles of PCl5, R is the ideal gas constant (0.0821 L atm/mol K) and T is the temperature (600 K). P = (nRT)/V = (0.01177 * 0.0821 * 600) / 0.5 = 11.562 atm (answer for a.)
03

Set up the ICE table and solve for equilibrium partial pressures

We know, PCl5(g) ⇌ PCl3(g) + Cl2(g) Initial moles: PCl5 = 0.01177 PCl3 = 0 Cl2 = 0 Change moles at equilibrium: PCl5 = -x PCl3 = x Cl2 = x Equilibrium moles: PCl5 = 0.01177 - x PCl3 = x Cl2 = x Partial pressures at equilibrium (dividing by volume): P_PCl5 = (0.01177 - x) / 0.5 P_PCl3 = x / 0.5 P_Cl2 = x / 0.5 Now, we know Kp = 11.5, and Kp = (P_PCl3 * P_Cl2) / P_PCl5 Solve for x: 11.5 = (x * x) / ((0.01177 - x)(0.5))
04

Calculate equilibrium partial pressures and mole fractions

Solve the equation above for x, which gives x = 0.00237 Equilibrium partial pressures are: P_PCl5 = (0.01177 - 0.00237) / 0.5 = 0.0188 atm (answer for b.) P_PCl3 = (0.00237) / 0.5 = 0.00474 atm P_Cl2 = (0.00237) / 0.5 = 0.00474 atm Total pressure at equilibrium is the sum of the partial pressures: P_total = P_PCl5 + P_PCl3 + P_Cl2 = 0.0188 + 0.00474 + 0.00474 = 0.02828 atm (answer for c.)
05

Calculate the percent dissociation at equilibrium

Percent dissociation is the ratio of moles dissociated to initial moles, multiplied by 100. Percent dissociation = (x / 0.01177) * 100 = (0.00237 / 0.01177) * 100 = 20.13% (answer for d.)

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Most popular questions from this chapter

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