Ethylene oxide is produced from the reaction of ethylene and oxygen at 280 C usi

ID: 526903 • Letter: E

Question

Ethylene oxide is produced from the reaction of ethylene and oxygen at 280 C using a silver catalyst.

2 C2H4 + O2= 2 C2H4O

40.0g of ethylene (MM= 28.054 g/mol) is mixed with 1.50 mol O2 in a 4.00 L reaction vessel at an initial temperature of 25.0 C.

a. Calculate the partial pressures of ethylene and oxygen in the reaction vessel at 25.0 C. Hint: no reaction occurs at this temperature.

b. Calculate the total pressure in the reaction vessel at 25.0 C.

c. The temperature is increased to 280. C to initiate the reaction. Calculate the theoretical yield of C2H4O.

d. The limiting reagent is ______.

e. Calculate the pressure of ethylene oxide (280.C) after the reaction is complete.

(a) Assuming ideal behavior,

Partial pressure of ethylene, PC2H4 = n R T / V = (40.0/28)*0.0821*298/4.00 = 8.74 atm

Partial pressure of oxygen, PO2 = 1.50 * 0.0821 * 298 / 4.00 = 9.17 atm

(b) Total pressure = 8.74 + 9.17 = 17.9 atm

(c)

Moles of ethylene = 40.0 / 28.0 = 1.43 mol

Moles of oxygen = 1.50 mol

From the balanced equation,

2 mol of ethylene needs 1 mol O2,

then, 1.43 mol of ethylene needs 1.43 /2 = 0.715 mol of O2 ( < 1.50 mol)

So, ethylene is limitinge reagent,

From the balanced equation,

2 mol of ethylene forms 2 mol ethylene oxide

then, 1.43 mol of ethylene forms 1.43 mol of ethylene oxide.

THerefore, mass of ethylene oxide (theoretical yield) formed =moles*molar mass=1.43*44=62.9 g.

(d) Thelimiting reagent is ethylene

(e)

Partial pressure ethylene oxide, PC2H4O = n R T / V = 1.43 * 0.0821 * (273+280) / 4.00 = 16.2 atm

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