CO(g) H2O(g)CO2(g) H2(g) Kp 0 0611 at 2000 K A reaction mixture initia...

For the reaction COgH2OgCO2gH2g Kp 00611 P CO2 P H2 P C ....

CO(g)+H2O(g)â‡ŒCO2(g)+H2(g) Kp=0.0611 at 2000 K A reaction mixture initially contains a CO partial pressure of 1360 torr and...

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Chemistry

CO(g)+H2O(g)â‡ŒCO2(g)+H2(g)
Kp=0.0611 at 2000 K
A reaction mixture initially contains a CO partial pressure of 1360torr and a H2O partial pressure of 1768 torr at 2000K.
Part A
Calculate the equilibrium partial pressure of CO2 and H2
? torr

Answer & Explanation Solved by verified expert

4.0 Ratings (659 Votes)

For the reaction CO(g)+H2O(g)â‡ŒCO2(g)+H2(g),

Kp = 0.0611 = P_CO2 . P_H2 /P_CO . P_H2O

We see the changes in pressure in of the species, as th reaction proceeds

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â CO(g)Â Â Â +Â Â Â H2O(g)Â Â Â â‡ŒÂ Â CO2(g)Â Â Â +Â Â Â Â H2(g)

Initial partial pressure (torr) Â Â Â Â Â Â Â Â Â Â 1360Â Â Â Â Â Â Â Â Â Â Â 1768Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

ChangeÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

At equilibriumÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (1360-x)Â Â Â Â Â Â Â Â Â (1768-x)Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Put equilibrium partial pressures in expression of Kp

Kp = 0.0611 = x.x / (1360-x). (1768-x)

x² = 0.0611 (1360-x). (1768-x)

x² = 0.0611 (2404480 -1360x -1768x + x²)

x² = (146913.728 -83.096x -108.025x +0.06112x²)

0.9389x² + 191.121x - 146913.728 = 0

Solve the quadratic equation, find the value of x.

x = 306 torr

P_CO2 = P_H2 = x = 306 torr

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