Question

For the reaction: CH₃OH(g) ↔ CO(g) + 2H₂(g), with the equilibrium concentrations for [CH₃OH]=0.20M, [CO]=0.44M, and [H₂]=2.7M, determine the gas phase equilibrium constant (K_p) at 400 ⁰C?

Answer 1

CH₃OH(g) ↔ CO(g) + 2H₂(g)

equilibrium constant=kc=[H2]^2 [CO]/[CH3OH]=(2.7M)^2 (0.44M)/(0.20M)=16.038 M^2

Kp=p(H2)^2* p(CO)/p(CH3OH) where p indicates the partial pressure of gases,p=cRT ,where c=concentration of the gases=[gas]

Kp={[H2]^2 (RT)^2}*{[CO]*RT}/{[CH3OH]*RT}=[H2]^2 [CO]/[CH3OH] *(RT)^(2+1-1)=Kc*(RT)^2

[in short Kp and Kc are related by the equation: Kp=Kc(RT)^n ,n=mol of product -mol of reactants]

kp=kc*(RT)^2 ,R=ideal gas constant=0.0821L atm/Kmol,T=400+273=673K

Kp=(16.038M^2 )((0.0821L atm/Kmol)*673K)^2=48962.846 L^2 atm^2   

Units: [(mol/L) *(L atm/Kmol)*K]^2=[(L atm)]^2

Kp=48962.846