Question

11.

Answer 1

Answer:

(1) -12 Kcal/mol

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!