A gas 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.

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meerkat18 meerkat18 · Answer 1 · 2017-11-02T13:10:59+01:00

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