Oxidation of methane
1 answer:
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Answer:
The carbocation intermediate reacts with a nucleophile to form the addition product.
Explanation:
The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.
The general mechanism is shown in the figure.
i) Attack of the electrophile on the benzene (which is the nucleophile)
ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.
iii) the carbocation formation is the rate determining step.
iv) There is no formation of addition product.
Thus the wrong statement is
The carbocation intermediate reacts with a nucleophile to form the addition product.
Answer:-
Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.
To calculate the heat absorbed or released we use the formula:
q = heat absorbed or released
m = mass of solution
s = specific heat capacity
and = change in temperature
mass of solution = mass of solute + mass of solvent
mass of solution = 5.00 g + 100.0 g = 105.0 g
(note:- density of pure water is 1 g per mL so the mass is same as its volume)
= 33.0 - 23.0 = 10.0 degree C
s =
Let's plug in the values in the formula and calculate q.
q =
q = 4389 J
To calculate the enthalpy of solution that is we convert q to kJ and divide by the moles of solute.
moles of solute =
= 0.118 moles
q = = 4.389 kJ
=
Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.
So,