Question

Ethylene (CH2CH2) is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 1010kg 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 kineticallycomplex reaction that produces ethylene gas and hydrogen gas. Suppose an engineer studying ethane cracking fills a 30.0L reaction tank with 24.0atm of ethane gas and raises the temperature to 800.°C. He believes Kp= 0.040 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.

Answer 1

Answer: The percent by mass of ethylene in the equilibrium gas mixture is 3.76 %

Explanation:

We are given:

Initial partial pressure or ethane = 24.0 atm

The chemical equation for the dehydration of ethane follows:

                   $C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)$

Initial:          24.0

At eqllm:    24-x            x              x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}$

We are given:

$K_p=0.040$

Putting values in above expression, we get:

$0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1$

Neglecting the value of x = -1 because partial pressure cannot be negative.

So, partial pressure of hydrogen gas at equilibrium = x = 0.96 atm

Partial pressure of ethylene gas at equilibrium = x = 0.96 atm

Partial pressure of ethane gas at equilibrium = (24-x) = (24 - 0.96) atm = 23.04 atm

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

$PV=nRT$          .........(1)

To calculate the mass of a substance, we use the equation:

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

• For ethane gas:

We are given:

$P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$23.04atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{23.04\times 30.0}{0.0821\times 1073}=7.85mol$

We know that:

Molar mass of ethane gas = 30 g/mol

Putting values in equation 2, we get:

$7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g$

• For ethylene gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of ethylene gas = 28 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g$

• For hydrogen gas:

We are given:

$P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$0.96atm\times 30.0L=n\times 0.0821\text{ L. atm }mol^{-1}K^{-1}\times 1073K\\\\n=\frac{0.96\times 30.0}{0.0821\times 1073}=0.33mol$

We know that:

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 2, we get:

$0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g$

To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:

$\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100$

Mass of equilibrium gas mixture = [235.5 + 9.24 + 0.66] = 245.4 g

Mass of ethylene gas = 9.24 g

Putting values in above equation, we get:

$\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%$

Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %