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At 1000 K, Kp = 19.9 for the reaction Fe2O3(s)+3CO(g)?2Fe(s)+3CO2(g) What are the equilibrium partial pressures...

At 1000 K, Kp = 19.9 for the reaction Fe2O3(s)+3CO(g)?2Fe(s)+3CO2(g)

What are the equilibrium partial pressures of CO and CO2 if CO is the only gas present initially, at a partial pressure of 0.904atm ?

Enter your answers numerically separated by a comma.

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Answer #1

Fe2O3(s) + 3CO(g) <---> 2Fe(s)+ 3CO2(g)

Initial -------- 0,904atm ------ 0atm

Change ------- -X ------- +X

Equilibrium -------- 0,904-Xat ------ X

Kp= (PCO2)3/ (PCO)3 (Solids and liquids are not affected by pressure)

==> 9,19= X3/(0,904-X)3 (replace the patrial pressures in equilibrium and solve for X)

==> ?9,19 = X/0,904-X

==>2,7098(0,904-x) = X

==>2,4497 -3,7098X = 0

==>-3,7098X = -2,4497

==>X=-2,4497/-3,7098

==> X=0,6603

Pressure of CO in eqiulibrium =0,904-0,6603 ==> 0,2437atm

Pressure of CO2 in equilibrium = X ==> 0,6603atm

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