I obtained 0.4008 atm which rounds to 0.401 atm. Did you convert the two pressures listed in torr to atm? That is a common student error.
1334/760 = ??
1778/760 = ??
Then
0.0611 = pCO2*pH2/pCO*pH2O
x*x/(1.755-x)(2.339-x)
Solve for x.
CO+H2O <--> CO2+H2
kp=0.0611 atm @2000K
A reaction mixture initially contains a CO partial pressure of 1334 torr and H2O partial pressure of 1778 torr at 2000K.
Whats the partial pressure of CO2 and H2?
I got 381 torr for both, but its not correct.
4 answers
132
304
It's actually 310 torr :)