Agas mixture of 50% co, 25% co2, and 25% h2 (by volume) is fed to a furnace at 900°c. determine the composition of the equilibrium co–co2–h2– h2o gas if the total pressure of the gas in the furnace is 1 atm.

We have to take some data like the energy needed for the formation of CO2, H2O, CO

We know that Go = H - TS

1kJ/mol = 238.846 cal/mol

C + 1/2O2 > CO                    Go1= -26700 - 20.95 T cal/mol

H2 + 1/2 O2 > H2O              Go2=-58900 + 13.1 T cal/mol

C + O2 > CO2                 Go3= -94200 - 0.2 T cal/mol

Now the reaction gives

                     H2   + CO2 > H2O + CO

Now Go4 = -8600 - 7.65 T cal/mol

At T( K )= 900oC = 900 + 273 = 1173 K , Go4= -8693 +7.65 X 1173 = 373.45 cal

Go4 = -RT ln K

ln K = (-373.45/ -(1.986 X 1173))

K = e0.160 = 1.173

                 H2      +       CO2 >        H2O +    CO

intial mole   0.25       0.25 >        X   +     0.5

After reaction (0.25 -X)     (0.25-X)             X          (0.5+X)

Now calculate for X, we know that K = product / reactant

K = (0.5+ X)* X / (0.25-X ) * (0.25-X)                     now K= 1.17

So,    1.173(0.0625- 0.5X-X2) = 0.5X- X2

0.0733- 1.0865X+ 0.173X2= 0

Calculate the value of X using quadratic equation

value of X = 6.81 % =0.068

So P(H2O)= 0.068

Total pressure = P(CO) + P(CO2) + P(H2) + P(H2O)=1

Now putting the value of X in the following

P(H2) = P(CO2)= 0.25- 0.068= 0.182

P(CO) = 0.5- 0.068= 0.568

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error