Respuesta :
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:
[tex]C_2H_6(g)\rightleftharpoons C_2H_4(g)+H_2(g)[/tex]
Initial: 24.0
At eqllm: 24-x x x
The expression of [tex]K_p[/tex] for above equation follows:
[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]
We are given:
[tex]K_p=0.040[/tex]
Putting values in above expression, we get:
[tex]0.040=\frac{x\times x}{24-x}\\\\x^2+0.04x-0.96=0\\\\x=0.96,-1[/tex]
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:
[tex]PV=nRT[/tex] .........(1)
To calculate the mass of a substance, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ..........(2)
- For ethane gas:
We are given:
[tex]P=23.04atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
Putting values in equation 1, we get:
[tex]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[/tex]
We know that:
Molar mass of ethane gas = 30 g/mol
Putting values in equation 2, we get:
[tex]7.85mol=\frac{\text{Mass of ethane gas}}{30g/mol}\\\\\text{Mass of ethane gas}=(7.85mol\times 30g/mol)=235.5g[/tex]
- For ethylene gas:
We are given:
[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
Putting values in equation 1, we get:
[tex]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[/tex]
We know that:
Molar mass of ethylene gas = 28 g/mol
Putting values in equation 2, we get:
[tex]0.33mol=\frac{\text{Mass of ethylene gas}}{28g/mol}\\\\\text{Mass of ethylene gas}=(0.33mol\times 28g/mol)=9.24g[/tex]
- For hydrogen gas:
We are given:
[tex]P=0.96atm\\V=30.0L\\T=800^oC=[800+273]K=1073K\\R=0.0821\text{ L. atm }mol^{-1}K^{-1}[/tex]
Putting values in equation 1, we get:
[tex]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[/tex]
We know that:
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 2, we get:
[tex]0.33mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=(0.33mol\times 2g/mol)=0.66g[/tex]
To calculate the mass percentage of ethylene in equilibrium gas mixture, we use the equation:
[tex]\text{Mass percent of ethylene gas}=\frac{\text{Mass of ethylene gas}}{\text{Mass of equilibrium gas mixture}}\times 100[/tex]
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:
[tex]\text{Mass percent of ethylene gas}=\frac{9.24g}{245.5g}\times 100=3.76\%[/tex]
Hence, the percent by mass of ethylene in the equilibrium gas mixture is 3.76 %
