No final temperature is given, so I assume it is negligible! Steam reforming of
ID: 1045317 • Letter: N
Question
No final temperature is given, so I assume it is negligible!
Steam reforming of methane (CH4) 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 1.5 L flask with 4.8 atm of methane gas and 3.1 atm of water vapor at 46.0 °C. He then raises the temperature, and when the mixture has come to equilibrium measures the partial pressure of carbon monoxide gas to be 2.2 atm. Calculate the pressure equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to 2 significant digits. K = 0x100 x n ?Explanation / Answer
no of mol of CH4 gas = PV/RT
= 4.8*1.5/(0.0821*319.15)
= 0.275 mol
No of mol of steam = 3.1*1.5/(0.0821*319.15)
= 0.177 mol
at equilbrium,
no of mol of CO = 2.2*1.5/(0.0821*319.15)
= 0.126 mol
CH4(g) + H2O(g) <-----> CO(g) + 3H2(g)
initial 0.275 0.177 - -
change -0.126 -0.126 +0.126 +0.378
equil 0.149 mol 0.051 mol 0.126 0.378 mol
total no of mol = 0.149+0.051+0.126+0.378 = 0.704 mol
ptotal = nRT/V
= 0.704*0.0821*319.5/1.5
= 12.3 atm
at equilibrium,
pCH4 = 0.149/0.704*12.3 = 2.6 atm
pH2O = 0.051/0.704*12.3 = 0.89 atm
pCO = 0.126/0.704*12.3 = 2.2 atm
pH2 = 0.378/0.704*12.3 = 6.6 atm
Kp = pCO*pH2^3/pCH4*pH2O
= 2.2*6.6^3/(0.89*2.6)
= 273.33
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