Question

Ethylene () is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 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 () 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 reaction tank with of ethane gas and raises the temperature to . He believes at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture. Round your answer to significant digits. Note for advanced students: the engineer may be mistaken about the correct value of , and the mass percent of ethylene you calculate may not be what he actually observes.

Answer 1

The question is incomplete, here is the complete question:

Ethylene is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10¹⁰ 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 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°C. He believes Kp = 0.4 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture.

Answer: The mass percent of ethylene gas is 9.20 %

Step-by-step explanation:

We are given:

Initial partial pressure of ethane gas = 38.0 atm

The chemical equation for the dehydrogenation of ethane follows:

`C_2H_6\rightleftharpoons C_2H_4+H_2`

Initial: 38

At eqllm: 38-x x x

The expression of
`K_p` for above equation follows:

`K_p=(p_(C_2H_4)* p_(H_2))/(p_(C_2H_6))`

We are given:

`K_p=0.40`

Putting values in above equation, we get:

`0.40=(x* x)/(38-x)\\\\x=-4.10,3.70`

Neglecting the negative value of 'x' because partial pressure cannot be negative

So, equilibrium partial pressure of ethane = 38 - x = 38 - 3.70 = 34.30 atm

Equilibrium partial pressure of ethene = x = 3.70 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

`PV=nRT` ..........(1)

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(2)

• For ethane:

We are given:

`P=34.3atm\\V=30.0L\\R=0.0821\text{ L atm }mol^(-1)K^(-1)\\T=400^oC=[400+273]=673K`

Putting values in equation 1, we get:

`34.3atm* 30.0L=n* 0.0821\text{ L atm }mol^(-1)K^(-1)* 673K\\\\n=(34.3* 30.0)/(0.0821* 673)=18.62mol`

Molar mass of ethane gas = 30 g/mol

Moles of ethane gas = 18.62 mol

Putting values in equation 2, we get:

`18.62mol=\frac{\text{Mass of ethane}}{30g/mol}\\\\\text{Mass of ethane gas}=(18.62mol* 30g/mol)=558.6g`

• For ethylene:

We are given:

`P=3.70atm\\V=30.0L\\R=0.0821\text{ L atm }mol^(-1)K^(-1)\\T=400^oC=[400+273]=673K`

Putting values in equation 1, we get:

`3.70atm* 30.0L=n* 0.0821\text{ L atm }mol^(-1)K^(-1)* 673K\\\\n=(3.70* 30.0)/(0.0821* 673)=2.01mol`

Molar mass of ethylene gas = 28 g/mol

Moles of ethylene gas = 2.01 mol

Putting values in equation 2, we get:

`2.01mol=\frac{\text{Mass of ethylene}}{28g/mol}\\\\\text{Mass of ethylene gas}=(2.01mol* 28g/mol)=56.28g`

• To calculate the mass percentage of substance in mixture, we use the equation:

`\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}* 100`

Mass of ethylene = 56.28 g

Mass of mixture = [558.6 + 56.28] g = 641.88 g

Putting values in above equation, we get:

`\text{Mass percent of ethylene}=(56.28g)/(641.88g)* 100=9.20\%`

Hence, the mass percent of ethylene gas is 9.20 %