Answer:
Answer is option C
Explanation:
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In Williamson ether synthesis, the reaction begins when the hydrogen from the alcohol's hydroxyl group contacts the halogen ion. Since the iodide ion is larger than the chloride ion, there is a larger chance of collision between it and the hydrogen atom, so butyl iodide is preferred over butyl chloride.
Also, Williamson synthesis is carried out at relatively high temperatures, around 50ºC-100º C. Ethyl iodide has a boiling point of 72.2ºC, so it will be in the liquid phase. Ethyl chloride has a boiling point of 12.3ºC so it will be in the gaseous phase, which is undesirable for this.
Ratio of neutrons to protons and <span>number of electrons in the outer shell</span>
There are there elements are important in organic compounds
carbon
hydrogen
oxygen :/example
glucose c6h12o6
Answer:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.025M)
Explanation:
Hello.
In this case, for this equilibrium problem, we first realize that at the beginning there is just HI, it means that the reaction should be rewritten as follows:

Whereas the law of mass action (equilibrium expression) is:
![Kc=\frac{[H_2][I_2]}{[HI]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH_2%5D%5BI_2%5D%7D%7B%5BHI%5D%5E2%7D)
That in terms of initial concentrations and reaction extent or change
turns out:
![Kc=\frac{x*x}{([HI]_0-2x)^2}\\\\54.3=\frac{x^2}{(0.391M-2x)^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7Bx%2Ax%7D%7B%28%5BHI%5D_0-2x%29%5E2%7D%5C%5C%5C%5C54.3%3D%5Cfrac%7Bx%5E2%7D%7B%280.391M-2x%29%5E2%7D)
And the solution via solver or quadratic equation is:

Whereas the correct answer is 0.183 M since the other value yield a negative concentration of HI at equilibrium (0.391-2*0.210=-0.029M).This, the equilibrium concentrations are:
![[H_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BH_2%5D_%7Beq%7D%3D0.183M)
![[I_2]_{eq}=0.183M](https://tex.z-dn.net/?f=%5BI_2%5D_%7Beq%7D%3D0.183M)
![[HI]_{eq}=0.391M-2*0.183M=0.025M](https://tex.z-dn.net/?f=%5BHI%5D_%7Beq%7D%3D0.391M-2%2A0.183M%3D0.025M)
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