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How do you know the difference between the enthalpy change and bond energy? Let's see the following questions. You don't need to calculate, but just see the formula.
1. Calculate the bond energy of a BrF bond using the following reaction equation. Delta H(f) of F5=-429, delta H(f) of Br=112, delta H(f) of F=79.
Br + 5F -> BrF5
They asked the bond energy, so we have to use the formula, delta H(reactant) - delta H (product). Then, we have to devide that answer by 5 because the question asks one B-F bond out of 5 B-F bond. However, Kaplan book says the formula used here is delta H (product) - delta H (reactant).
2.
CO (triple bond) 1075
C=O 728
C-Cl 326
Cl-Cl 243
Calculate the heat of reaction for the following equation using the information given above.
CO + Cl2 -> COCl2
They don't ask bond energy, so we have to use the formula, delta H (product) - delta H (reactant). However, Kaplan book says we have to use the formula, delta H(reactant) - delta H (product) in #2, so the answer is -62. Does it make sense? I think the answer of #2 is +62, not -62
1. Calculate the bond energy of a BrF bond using the following reaction equation. Delta H(f) of F5=-429, delta H(f) of Br=112, delta H(f) of F=79.
Br + 5F -> BrF5
They asked the bond energy, so we have to use the formula, delta H(reactant) - delta H (product). Then, we have to devide that answer by 5 because the question asks one B-F bond out of 5 B-F bond. However, Kaplan book says the formula used here is delta H (product) - delta H (reactant).
2.
CO (triple bond) 1075
C=O 728
C-Cl 326
Cl-Cl 243
Calculate the heat of reaction for the following equation using the information given above.
CO + Cl2 -> COCl2
They don't ask bond energy, so we have to use the formula, delta H (product) - delta H (reactant). However, Kaplan book says we have to use the formula, delta H(reactant) - delta H (product) in #2, so the answer is -62. Does it make sense? I think the answer of #2 is +62, not -62
