Question

A sample of gaseous PCl₅ was introduced into an evacuated flask so that the pressure of pure PCl₅ would be 0.50 atm at 523 K. However, PCl₅ decomposes to gaseous PCl₃ and Cl₂, and the actual pressure in the flask was found to be 0.74 atm. Calculate $K_p$ for the decomposition reaction PCl₅(g) $\rightarrow$ PCl₃(g) + Cl₂(g) at 523 K. Also, calculate K at this temperature.

Answer 1

Answer:

Kp = 0.2215

Kc = 0.0052

Explanation:

Step 1: The balanced equation

PCl5(g) → PCl3(g) + Cl2(g)

Step 2: given data

Initial pressure PCl5 = 0.50 atm

Temperature = 523 Kelvin

The actual pressure in the flask was found to be 0.74 atm.

Step 3: Calculate the pressure after the decomposition

Pressure of PCl5 after decomposition = 0.5 - X

Since the equation has a ratio 1:1 the pressure after the decomposition of PCl3 and Cl2 is X

Step 4: Calculate total pressure at equilibrium

0.74 atm = 0.5 atm - X + 2X

X = 0.74 - 0.5 = 0.24 atm

Step 5: Calculate Kp

Kp = P(PCl3)* P(Cl2) / P(PCl5)

Kp = 0.24² / 0.26 = 0.2215

Step 6: Calculate Kc

The ideal gas  wet: P*V = nRT

with P= the pressure of the gas in atm

with V= the volume of the gas in L

with n= the number of moles

with R= the gas constant in L*atm/mol*K

T = the temperature in Kelvin

P= nRT/V

Kp= Kc*(RT)^Δn

Kc=Kp/(RT)^Δn

Kc= 0.2215 / (0.08206*523)^1

Kc= 0.0052