Answer to Question #186187 in Chemistry for Kyle

Question

The two common chlorides of phosphorus, PCl3 and PCl5, both important in the production of other phosphorus compounds, coexist in equilibrium through the reaction PCl3(g) + Cl2(g) → PCl5(g).

At 250oC, an equilibrium mixture in a 2.50 L flask contains 0.105 g PCl5, 0.220 g PCl3, ande 2.12 g Cl2. What are the values of (a) Kc and (b) Kp for this reaction at 250oC? Write the full computation of your answer.

Answer 1

Solution:

The molar mass of PCl₅ is 208.24 g/mol.

The molar mass of PCl₃ is 137.33 g/mol.

The molar mass of Cl₂ is 70.906 g/mol.

Convert grams to Molarity (mol/L):

Molarity of PCl₅ = (0.105 g PCl₅ / 2.50 L) × (1 mol PCl₅ / 208.24 g PCl₅) = 0.0002017 M

[PCl₅] = 0.0002017 M

Molarity of PCl₃ = (0.220 g PCl₃ / 2.50 L) × (1 mol PCl₃ / 137.33 g PCl₃) = 0.0006408 M

[PCl₃] = 0.0006408 M

Molarity of Cl₂ = (2.120 g Cl₂ / 2.50 L) × (1 mol Cl₂ / 70.906 g Cl₂) = 0.0119595 M

[Cl₂] = 0.0119595 M

Write out the reaction:

PCl₃(g) + Cl₂(g) → PCl₅(g)

Write out the expression for K_c:

Plug in the Molarity given:

K_c = 26.319 M^-1

K_p = K_c × (RT)^Δn

Δn = 1 - (1+1) = -1

T = 250^oC + 273.15 = 523.15 K

R = 0.08206 L atm mol^-1 K^-1

Hence,

K_p = (26.319) × (0.08206 × 523.15)^-1 = 0.61307 = 0.613

K_p = 0.613 atm^-1

Answers:

(a) K_c = 26.319 M^-1

(b) K_p = 0.613 atm^-1