Answer to Question #215461 in General Chemistry for Mar

Question #215461

"Synthesis gas" is a mixture of carbon monoxide and water vapor. At high temperature synthesis gas will form carbon dioxide and hydrogen, and in fact this reaction is one of the ways hydrogen is made industrially. A chemical engineer studying this reaction fills a 500.mL flask with 2.0atm of carbon monoxide gas and 2.6atm of water vapor. When the mixture has come to equilibrium he determines that it contains 1.12atm of carbon monoxide gas, 1.72atm of water vapor and 0.88atm of hydrogen gas. The engineer then adds another 1.3atm of water, and allows the mixture to come to equilibrium again. Calculate the pressure of carbon dioxide after equilibrium is reached the second time. Round your answer to 2 significant digits.


1
Expert's answer
2021-07-09T05:15:51-0400

Lets start by drawing the ICE chart


reaction is


...............CO + H2O ========= CO2 + H2


I.............2.6.........3.4..........................0................0


E...........1.88........2.68......................0.72............0.72


since reaction is 1 : 1 if 0.72 atm of H2 are produced then the same amount of CO2 will be formed


calculate Kp this is the relationship between the pressure of products and the pressure of reactants


Kp = Pproducts / Preactants = [Pco2* PH2] / [PCO * PH2O]


Kp = 0.72 * 0.72 / (1.88 * 2.68) = 0.1029


Write the ICE chart again


...............CO + H2O ========= CO2 + H2


I.............1.88.........2.68+1.1.................0.72................0.72


C...........-x............. -x..............................+x........................+x

E......1.88 - x...........3.78-x.....................0.72+x..... ..0.72+x

Kp is = (0.72+x)*(0.72+x) /[ (1.88-x)*(3.78-x) ] = 0.1029

lets say that x is the change of pressure.

If you solve this equation you will get a value of x equal to 0.1

Pressure of CO2 is

0.72 + 0.1= 0.82 atm

using 2 sig figures this would be:8.2x10-1 atm


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS