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
Separate the components of a mixture.
Answer:
w = -531 kJ
1. Work was done by the system.
Explanation:
Step 1: Given data
- Heat gained by the system (q): 687 kJ (By convention, when the system absorbs heat, q > 0).
- Change in the internal energy of the system (ΔU°): 156 kJ
Step 2: Calculate the work done (w)
We will use the following expression.
ΔU° = q + w
w = ΔU° - q
w = 156 kJ - 687 kJ
w = -531 kJ
By convention, when w < 0, work is done by the system on the surroundings.
