The oxidation of ethylene to produce ethylene oxide proceeds according to 2C_2H_

Question

The oxidation of ethylene to produce ethylene oxide proceeds according to 2C_2H_4 + O_2 -rightarrow 2C_2H_4O The feed to a reactor contains 248 kmol/hr of ethylene and 167 kmol/hr oxygen. The conversion varies for the following 3 questions. If the reaction proceeds to completion: Find the % excess of the excess reactant. Calculate the molar flow rate of ethylene oxide in the product stream. If the reaction proceeds until the fractional conversion of the limiting reactant is 75%: Find the extent of reaction (kmol/hr). Calculate the component molar flow rates exiting the reactor. If the reaction proceed to a point where 90.8 kmol/hr of O_2 exit the reactor: Find the fractional conversion of ethylene and the fractional conversion of oxygen. Finally, calculate the component molar flow rates exiting the reactor.

Explanation / Answer

From the reaction,given above

2 moles ethylene needs 1 mole of oxygen gas to produce 2 mole ethylene oxide

Here, since the reactants are not according to the exact ratio as in equation , we use the concept of limiting agent

Here we find which reactant will end first and thus will decide the termination of the whole reaction as for a reaction to proceed all its reactant are needed to be present .

For this we divide number of moles of reactant to the factor they have been multiplied by in the equation.

a) Here no. of moles of ethylene =248(kmol/hr)

it,s factor in the equation =2

therefore for ethylene =248/2 =124(kmol/hr)

No. of moles of oxygen =167

it's factor in the equation =1

therefore, for oxygen =167/1 =167(kmol/hr)

since ethylene has less mole/(stoichiometric constant), therefore it is the limiting reactant, which means it will be used full or that there will be no ethylene left in equation.

Since. 2 mole of ethylene uses 1 mole of O2 , therefore 248 Kmole/hr will use 124 Kmoles/hr

Hence (167-124)Kmoles/hr of oxygen will be left =43.Kmole/hr

Also, molar flow rate of ethylene oxide is 248 Kmole/hr as 2 moles of ethylene produces 2 moles of ethylene oxide and ethylene is the limiting reactant here.

b) Since only 75% of limiting reactant is used , amount of ethylene used per hr =0.75*248

therefore extent of reaction is = change in amount of reactant/(stoichiometric constant) =(0.75*248-1*248)/2 = -31

Therefore, amount of ethylene left in the cylinder is .25*248 =62

SInce only 186 kmol/hr of ethylene is used in reactant , amount of oxygen used = 1/2* amount of ethylene consumed =93

Therefore amount of oxygen left =167-93 =74

Amount of ethylene oxide present =amount of ethylene used =186

c) Since 90.8 kmol/hr of oxygen exit the chamber,amount of oxygen used = 167-90.8=76.2 kmol/hr

Amount of ethylene used = 2* amount of oxygen used =2*76.2 =152.4

fractional conversion of ethylene used =0.6145

fractional convesion of oxygen = 0.45628

Amount of ethylene oxide coming out of chamber = amount of ethylene oxide used = 152.4Kmol/hr

Amount of ethylene exiting chamber =248-152.4 = 95.6 Kmol/hr

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