The molar enthalpy is 447 kJ and the reaction is endothermic
Answer:
How many distinct monochlorinated products can result when methylcyclohexane is subjected to free radical chlorination under UV light.
Explanation:
Consider 1-methylcyclohexane:
Its structure is shown below:
It has primary , secondary and tertiary carbons as shown in the image.
So, the following mono chlorinated product will be formed.
a) Ka for benzoic acid, C6H5CO2H <span>
HA + H2O ⇋ A^-
+ H3O^+ </span><span>
Ka = [A^-][H3O^+] / [HA]
C6H5CO2H + H2O ⇋
C6H5CO2^-^- + H3O^+ </span><span>
Ka = [C6H5CO2^-][H3O^+] / [C6H5CO2H] </span>
<span>
b) Kb for benzoate ion, (C6H3CO2)^ -
B + H2O ⇋
[BH]^+ + OH^- </span><span>
Kb = [[BH]^+][OH^-] / [B]
C6H3CO2^- + H2O ⇋
C6H5CO2H + OH^- </span><span>
Kb = [C6H5CO2H][OH^-] / [C6H3CO2^-] </span>
<span>
c) Kb for amiline, C6H5NH2 amiline? aniline
C6H5NH2 + H2O ⇋
[C6H5NH3]^+ + OH^- </span><span>
Kb = [[C6H5NH3]^+][OH^-] / [C6H5NH2] </span>
<span>
d)Ka for amilimium ion, (C6H5NH3)^+
C6H5NH3^+ + H2O ⇋
C6H5NH2 + H3O^+ </span><span>
<span>Ka = [C6H5NH2][H3O^+] / [C6H5NH3^+]</span></span>
|F| = 112 - 67 = 45 N
F = 45 N ( to the left)