Answer : The value of equilibrium constant K at 298 K is, 56.59
Explanation :
The given chemical reaction are:
(1) ;
(2) ;
First we have to determine the standard free-energy change for the following reaction.
(3) ;
Now we are reversing the reaction 1 and then adding reaction 1 and 2, we get:
(1) ;
(2) ;
Now we have to calculate the equilibrium constant K at 298 K.
where,
= standard Gibbs free energy = -10kJ/mol = -10000 J/mol
R = gas constant = 8.314 J/K.mol
T = temperature = 298 K
= equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant K at 298 K is, 56.59
Answer:
a) Structural formula for X and Y are attached below,
b) Systematic name of alcohol is Butan-2-ol.
c) X is secondary alcohol.
Explanation:
Alcohols are considered one of the most important class of organic compounds. They act as the starting materials in many important reactions like in synthesis of esters, ethers e.t.c.
Among different reactions shown by alcohols oxidation reactions are considered very important in formation of other intermediates like ketones, aldehydes and carboxylic acids.
Below are the oxidation reactions shown by different alcohols;
1) Primary Alcohol -------[O]--------> Aldehyde --------[O]--------> Carboxylic Acid
2) Secondary Alcohol --------[O]--------> Ketone -------[O]-------> No Reaction
3) Tertiary Alcohol ---------[O]---------> No reaction
Hence, oxidation takes place on alcohols which contain reactive C-H bonds. Therefore, in given scenario the alcohol is secondary alcohol and upon oxidation it produces a ketone.
Note:
In order to finalize the correct structure of final product we can assume that it is ketone because it is stated in statement that it cannot be oxidized further. Secondly the molar mass 72 g/mol was utilized to reach the final correct structure, for this purpose 28 amu (for CO carbonyl group) was subtracted from 72 and the remaining 44 amu was alloted to 2 CH₃ (Methyl part each having 15 amu mass) and CH₂ (Methylene part having 14 amu mass) respectively.
I was taught how to do it this way and trust me it is way easier than doing the stuff with fractions. I’ll upload a photo on how to do it
If the trend is linear, that is boiling point increases by a constant amount for each additional saturated carbon, the boiling point of octane would be 69*2 because you are adding two saturated carbons. So the predicted boiling point is 138C, which is very close to the literature value