Question

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.

Answer 1

• 1.4 atm²

Step-by-step 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:

`CH_4(g)+H_2O\rightarrow CO(g)+3H_2(g)`

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

Partial pressures in atm

`CH_4(g)+H_2O\rightarrow CO(g)+3H_2(g)`

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:

`K_p=((x)(3x)^3)/((0.60-x)(2.60-x))`

`K_p=((0.7667atm)(2.3atm)^3)/((3.9333atm)(1.7333atm))=1.37atm^2`

Round to two significant figures: 1.4 atm²