Question

The equilibrium constant, Kp, for the following reaction is 9.52×10-2 at 350 K: CH4(g) + CCl4(g) 2CH2Cl2(g) Calculate the equilibrium partial pressures of all species when CH4 and CCl4, each at an intitial partial pressure of 1.06 atm, are introduced into an evacuated vessel at 350 K. PCH4 = atm PCCl4 = atm PCH2Cl2 = atm

Answer 1

Answer:

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm

Explanation:

Step 1: Data given

The equilibrium constant, Kp= 9.52 * 10^-2

Temperature = 350 K

Each have an initial pressure of 1.06 atm

Step 2: The balanced equation

CH4(g) + CCl4(g) ⇆ 2CH2Cl2(g)

Step 3: The pressure at the equilibrium

pCH4 = 1.06 - X atm

pCCl4 = 1.06 - X atm

pCH2Cl2 = 2X

Step 4: Calculate Kp

Kp = (2X)² / (1.06 - X)*(1.06 - X)

9.52 * 10^-2 = 4X² / (1.06 - X)*(1.06 - X)

X = 0.1416

Step 5: Calculate the partial pressure

pCH4 = 1.06 - 0.1416 =  0.9184 atm

pCCl4 = 1.06 - 0.1416 =  0.9184 atm

pCH2Cl2 = 2 * 0.1416 = 0.2832 atm

Kp = (0.2832²) / (0.9184*0.9184)

Kp = 9.52 * 10^-2

pCH4 =  0.9184 atm

pCCl4 = 0.9184 atm

pCH2Cl2 = 0.2832 atm