O KINETICS AND EQUILIBRIUM Solving problems that mix equilibrium ideas with gas
ID: 563932 • Letter: O
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O KINETICS AND EQUILIBRIUM Solving problems that mix equilibrium ideas with gas laws 10 Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 × 10 kg of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane (CHCH3) from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas. artifial (CH,CH,) from natural g Suppose an engineer studying ethane cracking fills a 75.0 L reaction tank with 48.0 atm of ethane gas and raises the temperature to 450. °C. He believes Kn=0.10 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. Round your answer to 2 significant digits. Note for advanced students: the engineer may be mistaken about the correct value of Kp, and the mass percent of ethylene you calculate may not be what he actually observes.Explanation / Answer
The reaction for cracking ethane is
...................C2H6 ===== C2H4 + H2, draw the ICE chart
I...................48....................0.........0
C.................-x....................+x.........+x
E...............48-x..................x.............x
Kp is x*x/(48-x) = 0.1
if you solve this equation you will get a value of x equal to 2.142 atm
Pressure for Ethane is 48 - 2.142 = 45.858
Pressure for Ethylene is 2.142 atm
Pressure for Hydrogen is 2.142 atm
Total pressure is 50.142 atm (add all the partial pressures from before)
change these values from pressure to moles and then to mass using ideal gas equation
moles for ethane
moles = PV / RT = 45.858*75/(0.082*(450+273.15)= 58.008 moles
mass = moles * molar mass , molar mass for ethane is 30
mass = 58.008 * 30 = 17440.2 grams
repeat this for the hydrogen
moles of hydrogen =PV / RT = 2.142*75/(0.082*(450+273.15)= 2.709 moles
this is the same amount of ethylene
mass of hydrogen is = 5.41 grams
mass of ethylene is 76 grams
add all the masses
total mass = 1825.437 grams
percent mass of ethylene is (mass of ethylene divided by the total mass of the mixture)
76 / 1825.437 = 0.0416 *100 = 4.16 % using 2 significant figures is 4.2%
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