At 816 degree C and 1.00 atm pressure, a gaseous mixture or carbon monoxide and

Question

At 816 degree C and 1.00 atm pressure, a gaseous mixture or carbon monoxide and carbon dioxide is in equilibrium with solid carbon according to the reaction equation C (s) + CO_2 (g) 2 CO (g) If the gaseous mixture is 85.45% CO by mass, what is the equilibrium constant k_p (expressed in terms of the partial pressures in atm) at this temperature? (No units required.) For the above reaction, if the value of K_p (expressed in terms of the partial pressures in atm) was 0.201 at 638 degree C, what would be the value of K_eq (expressed in terms of the molar concentrations) at 638 degree C? (No units required.)

Explanation / Answer

Basis : 100 gm of gas mixture, Co=85.45 gm, moles of CO= mass/molar mass= 85.45/28 =3.02, mass of CO2=100-85.45= 14.55 gm, moles of CO2= 14.55/44 = 1.04

mole fraction = moles/total moles

Mole fractions : CO= 3.02/(3.02+1.04)= 0.74, CO2=1-0.74=0.26

Partial pressure : Mole fraction* total pressures

Partial pressures : CO= 1*0.74= 0.74 atm and CO2= 0.26*1= 0.26 atm

for the reaction C+CO2(g)----->2CO(g), Kp= [PCO2]/PCO2]=0.74*0.74/0.26= 2.11

Kp= KC*(RT)deltan, where deltan= moles of gas in the product- moles of gas in the reactant= 2-1= 1

Kp= KC*RT, at 638 deg,c T= 638+273= 911K

KC= KP/RT = 0.201/(0.0821*911) =0.0027

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