Steam forming of methane ( CH4) produces \"synthesis gas\" ,a mixture of carbon
ID: 1025900 • Letter: S
Question
Steam forming of methane ( CH4) produces "synthesis gas" ,a mixture of carbon monoxid gas and hydrogen gas, which is the starting. point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 100 L tank with 5.4 Mol of methan gas and 5.5 mol of water vapor at 40 degree celsius. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of hydrogen gas to be 13 mol. Calculate the concentration equilibrium constant for the steam reforming of methan at the final temperature of the mxiture. Round to 2 sig figs.
Explanation / Answer
The reaction is CH4(g)+ H2O(g)------->CO(g)+3H2(g)
molar ratio of CH4:H2O( Theoretical)= 1:1
actual molar ratio = 5.4:5.5 =1:1.02, so excess is H2O
let x= drop in moles of CH4 to reach equilibrium, at equilibrium, CH4= 5.4-x, H2O= 5.5-x and CO=x and H2= 3x
given 3x= 13, x=4.33, so at equilibrium, moles : CH4= 5.4-4.33= 1.07, CO=4.33, H2= 13 and H2O=5.5-4.33=1.17
concentration = moles/L, concentrations : CH4=1.07/100=0.0107M, CO= 4.33/100=0.0433M, H2= 13/100=0.13M and H2O= 1.17/100 =0.0117M
Kc= equilibrium constant =[H2]3[CO]/ [CH4][H2O]=0.13*0.13*0.13*0.0433/(0.0107*0.0117)=0.76