\"Synthesis gas\" is a mixture of carbon monoxide and water vapor. At high tempe

Question

"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 at 18.°C with 4.9atm of carbon monoxide gas and 0.83atm of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 4.62atm of carbon monoxide gas, 0.55atm of water vapor and 0.28atm of carbon dioxide. The engineer then adds another 2.5atm of carbon monoxide, and allows the mixture to come to equilibrium again. Calculate the pressure of hydrogen after equilibrium is reached the second time. Round your answer to 2 significant digits.

Explanation / Answer

CO. + H2O = CO2. + H2

At equilibrium 4.62 0.55 0.28 0.28

partial pressure of H2 is 0.28 atm since at equilibrium 0.28 atom is lost from the both the reactants i.e. 4.9-0.28= 4.62 and 0.83-0.28=0.55 atm.

Kp = 0.28 * 0.28 / (0.55 * 4.62) = 0.03

Let x be the amount of CO2 formed after the addition of 2.5 atms of CO

CO. + H2O = CO2 +. H2

New conc 4.62+2.5-x 0.55-x 0.28+x 0.28+x

Kp = (0.28+x)(0.28+x) / {(8.12-x)(0.55-x)}. or 0.03 = (0.28+x)(0.28+x) / {(8.12-x)(0.55-x)}

or 0.08 + .56x +x2 = 0.03 {4.47 -8.12x-0.55x +x2} or 0.08 + 0.56x + x2 = 0.13 -0.26x + 0.03x2

or (1-0.03)x2 + (0.56+0.26)x +0.08-0.13 =0 or 0.97x2 + 0.82x -0.05 =0

or x= 0.06

Partial pressure of H2 is 0.28 + 0.06 = 0.34 atm

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