The equilibrium constant is given as K=0.055 atm^-2 for the following reaction o

Question

The equilibrium constant is given as K=0.055 atm^-2 for the following reaction of formation of methanol from carbon monoxide and hydrogen in the gas- phase reaction. The mole fractions ("y"s) of species take place in the reaction at equilibrium satisfy the relation Y_CH_3OH/Y_CO Y^2_H_2 1/P^2 = K where P is the total pressure Write expressions for gram-moles of each species and mole fractions in terms of initial concentrations of the species and extent of reaction at equilibrium. If the reaction begins with 2 moles CO, 2 moles H_2 and no mole CH_3OH, determine the composition of the final product at equilibrium (both in moles and mole fractions) and the extent of reaction.

Explanation / Answer

Let x= moles of CO, y = moles of H2, z= moles of CH3OH

Let x’ =drop in moles of CO to reach equilibrium

At equilibrium

Moles : CO= (x-x’), H2= y-2x’ , CH3OH= z+x’

Total moles = x-x’+y-2x’+z+x’= x+y+z-2x’

Mole fractions = yCO= (x-x’)/ (x+y+z-2x’), yH2= (y-2x’)/ (x+y+z-2x’), yCH3OH= (z+x’)/ (x+y+z-2x’)

Partial pressures :   PCO =yCO*P, yH2= yH2*P, YCH3cOH= yCH3OH* P

Kp= PCH3OH/ (PCO* (PH2)2= YCH3OH/(YCO*Y2H2*P2)

2. given x=2 , y= 2, z=0

yCO= (2-x’)/(4-2x’), YH2= ( (2-2x’)/(4-2x’), YCH3OH= x’/(4-2x’)

given Kp=0.055

x’/(4-2x’)/ {(2-x’)/(4-2x’’)* [ 2-2x’)/(4-2x’)2 = (1.4)2*0.055 =0.1078

x’*(4-2x’)2/ [ (2-x’)* (2-2x’)2= 0.1078

x’*(2-x’)2/ (2-x’)*(1-x’)2= 0.1078

x’*(2-x’)/ (1-x’)2= 0.1078

when solved using excel, x’ =0.04984

extent of reaction 100*0.04984/2 = 2.492%

moles at equilibrium, CO= 2-0.04984 =1.9506, H2= 2-2*0.04984= 1.900, CH3OH= 0.04984

total moles = 1.9506+1.900+0.04984= 3.9

Mole fractions : CO= 1.9506/3.9=0.500, H2= 1.9/3.9= 0.48, CH3OH= 0.04984/3.9= 0.02

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.