If the mass percentage composition of a compound is 72.1% Mn and 27.9% O, its empirical formula is
1 answer:
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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, <u>two bonds are broken</u>, a C-H and a Br-Br. Additionally, a C-Br and a H-Br are <u>formed</u>.
If we want to calculate the enthalpy value, we can use the equation:
<u>ΔH=ΔHbonds broken-ΔHbonds formed</u>
If we use the energy values reported, its possible to calculate the energy for each set of bonds:
<u>ΔHbonds broken</u>
<u />
C-H = 94.5 Kcal/mol
Br-Br = 51.5 Kcal/mol
Therefore:
105 Kcal/mol + 53.5 Kcal/mol = 146 Kcal/mol
<u>ΔHbonds formed</u>
C-Br = 70.5 Kcal/mol
H-Br = 87.5 Kcal/mol
Therefore:
70.5 Kcal/mol + 87.5 Kcal/mol = 158 Kcal/mol
<u>ΔH of reaction</u>
<u />
ΔH=ΔHbonds broken-ΔHbonds formed=(146-158) Kcal/mol = -12 Kcal/mol
I hope it helps!
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