Answer:
(1) -12 Kcal/mol
Step-by-step explanation:
Our answer options for this question are:
(1) -12 Kcal/mol
(2) -13 Kcal/mol
(3) -15 Kcal/mol
(4) -16 Kcal/mol
With this in mind, we can start with the chemical reaction (Figure 1). In this reaction, two bonds are broken, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are formed.
If we want to calculate the enthalpy value, we can use the equation:
ΔH=ΔHbonds broken-ΔHbonds formed
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
ΔHbonds broken
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
ΔHbonds formed
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
ΔH of reaction
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it helps!