For the reaction CO+H20<--->H2+CO2, Keq=5.10 at 527 C. If the initial concentration [CO] = 0.15M, [H2O] = 0.25M, [H2] = 0.42M and [CO2] = 0.37M, how does the reaction proceed.
The chemical equation CO + H2O <---> H2 + CO2 has a Keq of 5.10 at 527 C.
The Keq for the reaction is defined as Keq = [H2][CO2]/[CO][H2O] = 5.10
The initial concentration of the compounds taking part in the reaction are given as [CO] = 0.15M, [H2O] = 0.25M, [H2] = 0.42M and [CO2] = 0.37M
(0.42*0.37)/(0.15*0.25) = 4.144
As this is less than Keq, the reaction could proceed forward leading to the formation of more H2 and CO2.
If moles of CO and H2O react to form x moles of H2 and CO2, we can estimate x by solving:
((0.42+x)*(0.37+x))/((0.15-x)*(0.25-x) = 5.1
The solution for x that is valid is 0.0129.
The final molarity of the gases is [CO]= 0.137, [H2O] = 0.237, [H2] = 0.4329 and [CO2] = 0.3829