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
