The solution is as follows:
K = [Partial pressure of isoborneol]/[Partial pressure of borneol] = 0.106
The molar mass of isoborneol/borneol is 154.25 g/mol
Mol isoborneol = 15 g/154.25 = 0.0972 mol
Mol borneol = 7.5 g/154.25 = 0.0486 mol
Use the ICE approach
borneol → isoborneol
I 0.0972 0.0486
C -x +x
E 0.0972 - x 0.0486 + x
Total moles = 0.1458
Using Raoult's Law,
Partial Pressure = Mole fraction*Total Pressure
[Partial pressure of isoborneol] = [(0.0972-x)/0.1458]*P
[Partial pressure of borneol] = [(0.0486+x/0.1458)]*P
0.106 = [(0.0972-x)/0.1458]*P/ [(0.0486+x/0.1458)]*P
Solving for x,
x = 0.0832
Thus,
<em>Mol fraction of borneol = (0.0486+0.0832)/0.1458 = 0.904</em>
<em>Mol fraction of isoborneol = (0.0972-0.0832)/0.1458 = 0.096</em>
A. BeCl2 sp2
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It follows that the reaction is spontaneous at high temperatures Option A.
<h3>What is ΔS ?</h3>
The term ΔS is referred to as the change in the entropy of the system. Now recall that entropy is defined as the degree of disorderliness in a system. If a system is highly disorderly then it means that it has a high entropy. Also, ΔH has to do with the heat change that accompanies a reaction.
We know that both the entropy and the heat change can both either be positive or negative. Now we know that the equation ΔG = ΔH - TΔS can be used to ascertain whether or not a reaction will be spontaneous. If the result is negative, then the reaction will be spontaneous.
As such, when then it follows that the reaction is spontaneous at high temperatures Option A.
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Answer:
Ether
SN1 mechanism
Explanation:
The nucleophile in this reaction is CH3OH. It is a poor nucleopile. We already know that a poor nucleophile reacting with a tertiary alkyl halide often leads to the substitution product as the major product.
Also, the iodide ion is a good leaving group. This makes the SN1 substitution more likely yielding the ether as the major product as shown in the image attached.
