Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
The methane molecule undergoes oxidation and the carbon is oxidized to carbon dioxide and hydrogen to water.
In this reaction there is cleavage of four C-H bonds and two O=O bonds
there will be formation of two C=O bonds and four O-H bonds.
Overall due to more bond cleavage energy there will be evolution of energy. thus combustion is an exothermic reaction.
Answer:
Explanation:
2C₂H₅OH = C₄H₆ + 2H₂O + H₂
2 mole 1 mole
molecular weight of ethyl alcohol
mol weight of C₂H₅OH = 46 gm
mol weight of C₄H₆ 54 gm
540 gm of C₄H₆ = 10 mole
10 mole of C₄H₆ will require 20 mol of ethyl alcohol .
20 mole of ethyl alcohol = 20 x 46
= 920 gm
ethyl alcohol required = 920 gm .
Boyle's law
I hope this helps
To separate off different products in order of their boiling points. You do it by a process of heating and cooling in a horizontal condenser usually.
