Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical
ID: 705091 • Letter: E
Question
Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 × 1010 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 (CH3CH3) from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas. Suppose an engineer studying ethane cracking fills a 30.0 L reaction tank with 38.0 atm of ethane gas and raises the temperature to 400·?. He believes Kp 0.40 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 0% x10Explanation / Answer
The reaction taking place is C2H6 --------->C2H4+ H2
given initial pressure of C2H6= 38 atm
let P= drop in pressure of C2H6 to reach equilibrium, At equilibrium, partial pressures : PC2H4= P, PH2=P and PC2H6=38-P
P denotes partial pressure of the component.
KP= PC2H4*PH2/(PC2H6)= P2/(38-P) =0.4, when solved using excel, P= 3.71 atm
partial pressure of ethylene at equilibrium= 38-3.71=34.29 atm
total pressure at equilibrium= 3.71 ( ethylene)+ 3.71 ( hydrogen)+ 34.29 ( ethane)= 41.71 atm
moles of each gas can be calculated using gas law and based on partial pressure of gas that is defined as the pressrue exerted if the gas alone occupies the entire volume, since partial pressure of both Hydrogen and C2H4 are same
moles of C2H4= moles of Hydrogen = PH2( or PC2H4)*V/RT ={3.71* 30/(0.0821*(400+273))}= 2.014 moles
mass of C2H4= moles* molar mass= 2.014* 28 =56.4 gm, mass of H2= moles of Hyrgoen* molar mass =2.014* 2= 4.028 gm,
moles of C2H6 = PC2H6*V/RT = 34.29*30/(0.0821*673)= 18.62 moles, mass of C2H6= moles* molar mass = 18.62*30 =558.6 gm
total mass of mixture at equilibrium= 56.4( C2H4)+4.028(hyrogen)+ 558.6 ( C2H6)= 619.028 gm
mass % of C2H4 in the sample = 100*56.4/619.028=9.11% at equilibrium,
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