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Steam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a flask with of methane gas and of water vapor at . She then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of carbon monoxide gas to be . Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

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Answer:

  • 1.4 atm²

Explanation:

The missing data on the question are:

  • . . . fills a 200. mililiter flask
  • . . . with 4.7 atm methane gas
  • and 2.5 atm of water
  • at 55.0ºC
  • partial pressure of cabon dioxide to be 2.3 atm

Solution

1. Equilibrium equation:

    [tex]CH_4(g)+H_2O\rightarrow CO(g)+3H_2(g)[/tex]

2. ICE (initial, change, equilibrium) table:

Partial pressures in atm

          [tex]CH_4(g)+H_2O\rightarrow CO(g)+3H_2(g)[/tex]

I           4.7            2.5           0            0

C          -x             -x          +x          +3x

E        4.7 - x     2.5  - x       x           3x

3. Partial pressures:

The partial pressure of hdyrogen gas at equilibrium is 2.3 atm

Then:

  • 3x = 2.3 atm
  •  x = 0.7667 atm
  • 4.7 - x = 3.9333 atm
  • 2.5 - x = 1.7333 atm

4. Pressure equilibrium constant equation:

      [tex]K_p=\dfrac{(x)(3x)^3}{(0.60-x)(2.60-x)}[/tex]

[tex]K_p=\dfrac{(0.7667atm)(2.3atm)^3}{(3.9333atm)(1.7333atm)}=1.37atm^2[/tex]

Round to two significant figures: 1.4 atm²