Question

Consider the water-gas shift reaction:

H2(g) + CO2(g) = H2O(g) + CO(g)

A) Find Kp for this reaction at 298 K

B) What is

1) Consider the water-gas shift reaction:

H2(g) + CO2(g) = H2O(g) + CO(g).

A) Find Kp for this reaction at 298 K.

B) What is ΔH° for this reaction?

C) What type of pressure and temperature change would favor CO production?

D) What is K at 1000 K?

Answer 1

CO2(g) + H2(g) --><-- H2O(g) + CO(g) : ΔG =28.52 Kj/mol

ΔG= -RT lnK

lnK= - 28.52*1000/(298*8.314)= -11.51

K= 1*10^-5

Enthalpy change = enthalpy of products- enthalpy of reactants

enthalpy of H2O+ enthalpy of CO- ( enthalpy of CO2+0) ( enthapy change of element =0

=-242+-110.5 -(-393.5)=41 Kj/mole

C) Since the reaction is endothermic, when temperature is increased, the reaction proceeds in the endothermic direction. Hence increase in temperature favors forward reaction.Pressure does not have any affect on the equilibrium of reaction.

d) ln(K2/K1)= (delH/R)*(1/T1-1/T2)

K2 equilibrium constant at 1000 K T2= 1000K T1= 298 K1= 1*10^-5 delH= 41Kj/mole= 41*1000 j/mole

ln(K2/K1)= (41*1000/8.314)*(1/298-1/1000)=11.61

K2/K1= 110970.4

K2= 110970.4*1*10^-5 =1.109