Question

No. 5

A 1.00L bulb containing 0.233 Mol N2 and 0.341 mol PCl5 is heated up to 523K. The total pressure at equilibrium is 29.33 bar. Assuming that all gases are ideal, calculate Kp for the only reaction that occurs: PCl5(g)<--->PCl3(g) +Cl2(g) (Hint: Notice that the total number of moles at equilibrium is not the same as the initial number of moles)

Answer 1

According to the ideal gas law pV = nRT   => n = pV/RT

29.33 bar x 0.987 atm/bar = 28.95 atm

n = (28.95 atm x 1 L) /(0.08206 L atm K⁻¹ mol⁻¹ x 523 K) = 0.6745 mol   The total number of moles at equilibrium.

Since N2 is not involved in the reaction, its quantity remains the same, so the amount of gases due to the PCl5 decomposition will be

0.6745 mol - 0.233 mol = 0.4415 moles

Since 1 mol of PCl5 decomposes to 1 mol of PCl3 and 1 mol of Cl2, the change in the amount of gasses that occurred at the equilibrium is equal to the number of moles PCl5 decomposed, so the new concentration of PCl5 will be

[PCl5] = (0.341 - (0.4415 - 0.341)) mol/1 L =(0.341 - 0.1005) mol / 1 L = 0.2405 M

The concentrations of PCl3 and 1 mol of Cl2 will be equal to

[PCl3] = [Cl2] = 0.1005 mol /1 L = 0.1005 M

PCl5(g)<--->PCl3(g) +Cl2(g)

Kp = [PCl3][Cl2]/[PCl5] = 0.1005²/0.2405 = 0.042 = 4.2 x 10^-2