DOLL CHEM 104 Homework 4 5. 20 points-Ch. 17.3, 17,5 Steam reforming of natural
ID: 564554 • Letter: D
Question
DOLL CHEM 104 Homework 4 5. 20 points-Ch. 17.3, 17,5 Steam reforming of natural gas is the method used to manufacture most of the hydrogen gas us in industry currently. Although natural gas is mostly composed of methane, it also usually other hydrocarbons, such as propane, C3He. Imagine the steam reforming of propane od contains two steps: Step 1: Chela, + 3 H:01g) #3 cout 7hhw Step 2: COH2Ou CO2+H2 K, = 8.175x 10s at 1200. K Kp 0.6944 at 1200. K a) Write the overall equation for the reaction of propane and steam to produce carbon dioxide and hydrogen. Calculate Kp and K, for the overall process at 1200. K. Suppose a reactor is filled with a mixture of propane and steam at 1200. K in which the partial pressure of C,Hg 1.00 atm and the partial pressure of H20 4.00 atm. After the reaction reaches equilibrium, what is the final total pressure in the reactor? (Hint: you may assume that all gases behave ideally. What assumption can you make about the extent of the reaction at equilibrium?) b) c)Explanation / Answer
Step 1:
C3H8(g)+ 3H2O<-> 3CO(g)+ 7H2(g), Kp=8.175*1015
Kp1 = [PCO]3 [PH2]7/ [PC3H8][PH2O]3= 8.175*1015 (1)
Step 2: CO+ H2O<---> CO2(g)+ H2(g), Kp= 0.6944
KP2=[PCO2][PH2]/[PCO] [PH2O]= 0.6944
=Kp2’= Kp23= [PCO2]3 [PH2]3/ [PCO]3 [PH2O]3 =(0.6944)3 = 0.3348 (2)
Eq.2* Eq.1 gives
Kp = [PCO2]3 [PH2]10 / [PH2O]6 [PC3H8] = 8.175*1015 *0.3348 =2.73*1015 (3)
The overall reaction is C3H8+ 6H2O <------->3CO2+10H2
Kp =KC*(RT)deltan, deltan= change in no of moles during the reaction = 10+3-(1+6)=6
2.73*1015= KC*(0.0821*1200)6, K= 2985.5
Given partial pressure of C3H8= 1 atm and that of H2O= 4 atm,
Let x= drop in pressure of C3H8 to reach equilibrium
C3H8(g)+ 3H2O<-> 3CO(g)+ 7H2(g), Kp=8.175*1015
Since Kp is high, the reaction is complete and since 1 atm of C3H8 and 4 atm of H2O are available. 3 atms of H2O participate in the 1st reaction giving rise to 3 atm of CO and 7 atm of H2. The 1 atm of H2O participates in the second reaction.
The second reaction is CO(g)+ H2O(g) <--->CO2(g)+ H2(g)
Let x= drop in pressure of H2O to reach equilibrium.
At equilibrium, CO= 3-x, H2O= 1-x and CO2= x and H2= 7+x
Kp= 0.6944= x*(7+x)/ (3-x)*(1-x)
When solved, x= 0.2117, So at equilibrium, PCO= 3-0.2117= 2.7883 atm, PH2O= 1-0.2117=0.7883 atm, PCO2=0.2117 atm and PH2= 7.217
Total pressure at equilibrium= 2.7883+0.7883+0.2117+7.217= 11 atm
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