Respuesta :
Answer: The number of moles of ethanol after equilibrium is reached the second time is 11. moles.
Explanation:
We are given:
Initial moles of ethene = 34 moles
Initial moles of water vapor = 15 moles
The chemical equation for the formation of ethanol follows:
[tex]CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH[/tex]
Initial: 34 15
At eqllm: 34-x 15-x x
We are given:
Equilibrium moles of ethene = 24 moles
Equilibrium moles of water vapor = 5 moles
Calculating for 'x'. we get:
[tex]34-x=24\\\\x=10[/tex]
Volume of container = 100.0 L
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[CH_3CH_2OH]}{[CH_2=CH_2][H_2O]}[/tex] .......(1)
[tex][CH_3CH_2OH]=\frac{10}{100}=0.1M[/tex]
[tex][CH_2=CH_2]=\frac{24}{100}=0.24M[/tex]
[tex][H_2O]=\frac{5}{100}=0.05M[/tex]
Putting values in expression 1, we get:
[tex]K_c=\frac{0.1}{0.24\times 0.05}\\\\K_c=8.3[/tex]
Now, 11 moles of ethene gas is again added and equilibrium is re-established, we get:
The chemical equation for the formation of ethanol follows:
[tex]CH_2=CH_2+H_2O\rightleftharpoons CH_3CH_2OH[/tex]
Initial: 24+11 5 10
At eqllm: 35-x 5-x 10+x
[tex][CH_3CH_2OH]=\frac{10+x}{100}[/tex]
[tex][CH_2=CH_2]=\frac{35-x}{100}[/tex]
[tex][H_2O]=\frac{5-x}{100}[/tex]
Putting values of in expression 1, we get:
[tex]8.3=\frac{\frac{(10+x)}{100}}{\frac{(35-x)}{100}\times \frac{(5-x)}{100}}\\\\8.3=\frac{(10+x)\times 100}{(35-x)\times (5-x)}\\\\x^2-52x+55=0\\\\x=50.9,1.1[/tex]
The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.
Moles of ethanol = [tex](10+x)=10+1.1=11.[/tex]
Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.
The number of moles of ethanol reached is : 11 moles
Given data :
moles of ethene ( initial ) = 34
moles of water vapor ( initial ) = 15
Volume of container = 100 L
Expressing the Chemical equation for the formation of ethanol
CH₂ + H₂O ⇄ CH₃CH₂OH ----- ( 1 )
at Equilibrium : 34-x 15-x x
Given that :
Equilibrium moles of ethene = 24
Eq moles of water vapor = 5
solve for x
x = 10
Following the expression of Kc for equation 1
The concentration of [ H₂O ] = 5 / 10 = 0.05 M
Kc = [tex]\frac{0.1 }{0.24 * 0.05}[/tex] = 8.3
Given that 11 moles of ethene is added again and equilibrium is reestablished
CH₂ + H₂O ⇄ CH₃CH₂OH
At equilibrium : 35-x 5-x 10+x
[ H₂O ] = ( 5-x ) / 100
solve for x using the expression for Kc
x = 50.9 , 1.1
we neglect the value of 50.9
therefore x = 1.1
moles of ethanol = 10 + 1.1 = 11
Hence we can conclude that the number of moles of ethanol after equilibrium is reached the second time is 10 + 1.1 = 11
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