Question

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2 and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium is established at 500 K...
Co2 (g) + H2 (g) image from custom entry tool CO (g) + H2O (g)
a. Calculate the initial partial pressures of CO2, H2, and H2O.
b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.
c. Calculate Kp for this reaction

Answer 1

Answer:

a) p(CO2) = 4.103 atm

p(H2) = 2.0515 atm

p(H2O) = 3.2824 atm

b) pCO2  = 3.8754 atm

pH2 =  1.8239 atm

pCO = 0.2276 atm

pH2O = 3.51 atm

c) Kp= 0.113

Explanation:

Step 1: Data given

Moles of CO2 = 0.2000 mol

Moles of H2 = 0.1000 mol

Moles of H2O = 0.1600 mol

Volume = 2.000 L

Temperature = 500 k

Step 2: The balanced equation

CO2 (g) + H2 (g) → CO (g) + H2O (g)

Step 3: Calculate the initial partial pressures of CO2, H2, and H2O.

pV = nRT

P = (nRT)/V

p(CO2) = (0.2000 mol * 0.08206 * 500)/2.000L

⇒ p(CO2) = 4.103 atm

p(H2) = (0.1000 mol * 0.08206 * 500)/2.000L

⇒ p(H2) = 2.0515 atm

p(H2O) = (0.1600 mol * 0.08206 * 500)/2.000L

⇒ p(H2O) = 3.2824 atm

b. At equilibrium PH2O= 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO.

Step 1: Calculate the change in pH2O

The change in pH2O = 3.51 - 3.2824 = 0.2276

Step 2: the initial pressures

pCO2 = 4.103 atm

pH2 = 2.0515 atm

pCO = 0 atm

pH2O = 3.2824

Step 3: The partial pressure at the equilibrium

Since there reacts 0.2276 atm for H2O, and the mol ratio is 1:1:1:1

For each gas, there will react 0.2276 atm

pCO2 = 4.103 - 0.2276 = 3.8754 atm

pH2 = 2.0515 - 0.2276 = 1.8239 atm

pCO = 0 + 0.2276 = 0.2276 atm

pH2O = 3.51 atm

c. Calculate Kp for this reaction

Kp = (pCO * pH2O)/ (pCO2 * pH2)

Kp = (0.2276 * 3.51) /( 3.8754 *1.8239)

Kp = 0.113