At a certain temperature, iron (II) oxide, FeO, can react with carbon monoxide, CO, to form elemental iron, Fe, and carbon dioxide, CO2. The value of Kp at that temperature is 0.242. What is the pressure of CO2 at equilibrium if a sample of FeO was initially in a container with CO at a pressure of 0.95 atm? The chemical reaction involved in this process is: FeO(s) + CO(g) ⟷ Fe(s) + CO2(

The pressure of CO2 at the equilibriumis 0.185 atm

Explanation:

Step 1: Data given

The value of Kp at that temperature is 0.242

FeO was initially in a container with CO at a pressure of 0.95 atm

Step 2: The balanceequation

FeO(s) + CO(g) ⟷ Fe(s) + CO2(g)

Step 3:The initial pressure

pCO = 0.95 atm

pCO2 = 0 atm

Step 4: The pressure at the equilibrium

pCO = 0.95 - x atm

pCO2 = X atm

Step 5: Calculate Kp

Kp = pCO2 / pCO

0.242 = X / (0.95 - X)

X = 0.185

pCO = 0.95 - 0.185 =  0.765 atm

pCO2 = 0.185 atm

The pressure of CO2 at the equilibriumis 0.185 atm

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :