<u>Answer:</u> The number of moles of ethanol after equilibrium is reached the second time is 11. moles.
<u>Explanation:</u>
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:

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:

Volume of container = 100.0 L
The expression of
for above equation follows:
.......(1)
![[CH_3CH_2OH]=\frac{10}{100}=0.1M](https://tex.z-dn.net/?f=%5BCH_3CH_2OH%5D%3D%5Cfrac%7B10%7D%7B100%7D%3D0.1M)
![[CH_2=CH_2]=\frac{24}{100}=0.24M](https://tex.z-dn.net/?f=%5BCH_2%3DCH_2%5D%3D%5Cfrac%7B24%7D%7B100%7D%3D0.24M)
![[H_2O]=\frac{5}{100}=0.05M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Cfrac%7B5%7D%7B100%7D%3D0.05M)
Putting values in expression 1, we get:

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:

Initial: 24+11 5 10
At eqllm: 35-x 5-x 10+x
![[CH_3CH_2OH]=\frac{10+x}{100}](https://tex.z-dn.net/?f=%5BCH_3CH_2OH%5D%3D%5Cfrac%7B10%2Bx%7D%7B100%7D)
![[CH_2=CH_2]=\frac{35-x}{100}](https://tex.z-dn.net/?f=%5BCH_2%3DCH_2%5D%3D%5Cfrac%7B35-x%7D%7B100%7D)
![[H_2O]=\frac{5-x}{100}](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Cfrac%7B5-x%7D%7B100%7D)
Putting values of in expression 1, we get:

The value of 'x' cannot exceed '35', so the numerical value of x = 50.9 is neglected.
Moles of ethanol = 
Hence, the number of moles of ethanol after equilibrium is reached the second time is 11. moles.