The halogens fluorine (F2) AND BROMINE (Br2)?
Bond F-F Br-Br
Molar bond enthalpy/KJ 159 193
The overall enthalpy change (∆H) for the reaction of the five moles of fluorine with one mole of bromine is -429KJ
Write down a balanced thermochemical equation for this reaction.
Answer:
The reaction is
5F2 + Br2 -----> 2BrF5
(159) (193) (∆H) (x)
Suppose x is the bond energy for F-Br bond.
There are 10 of these bonds being formed.
So, by Hess's law
∆H = Bond energy of products - Bond energy of reactants
∆H = (5 * 159 + 1 * 193) - 10x
-429 = 988 - 10x
So , 10x = (429+988)
x= 141.7 kJ/mol
The bond energy of F-Br is 141.7 kJ/mol
the equation would be
Br2 + 5F2 --- > 2BrF5
Delta H for the reaction = delta h of the products - delta H of the reactants.
Delta H of the reactants = 988 (193 + 5 (159)) kj
therefore Delta H of the products = -429+988 = 559 kj
Put the values of enthalpy changes in the equation and u have a balanced thermochemical equation in place.
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