Chem 2 Question

Question

Bromine trifluoride forms according to the following balanced chemical equation:
BrF(g)+F2(g)→BrF3(g)
Calculate the change in enthalpy for this reaction using the following known reaction enthalpies:
2BrF3(g)→Br2(l)+3F2(g) ΔH=511.2kJ
1/2Br2(l)+1/2F2(g)→BrF(g) ΔH=−93.8kJ

 

We have been stuck on this for a while now. Any help would be greatly appreciated. Thank you!

Answer 1

John,

 

The first reaction needs to reversed  and divided by 2 to generate 1 BrF₃ in the final equation.  When you divided the equation by 2 and reverse the reaction, the sign of ΔH will be reversed and divided by 2.  Therefore the first reaction will have a ΔH = -255.6kJ

 

The second reaction needs to be reversed to get BrF as reactant in the final equation.  When you reverse, the sign of ΔH changes.  Therefore, the second reaction will have a ΔH = +93.8kJ

 

If you had both of these reactions to generate the final balanced chemical equation, the reaction enthalpy will be as follows: ΔH = -255.6kJ + 93.8kJ = -161.8kJ

 

Hope it helps.