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Sagot :
Using standard heats of formation, the standard enthalpy change for the reaction is mathematically given as
= 542.11 kJ
What is the standard enthalpy change for the following reaction?
Generally, the equation for standard enthalpy change is mathematically given as
[tex]\Delta H rxn = \Delta H product - \Delta H reactant[/tex]
therefore,
[tex]\Delta H rxn = (\Delta H Br_2 + \Delta H 3F_2) - (\Delta H 2BrF_3)[/tex]
= (30.91 kJ + 3*0 kJ) - (2*(-255.6) kJ)
= 542.11 kJ
In conclusion, the standard enthalpy change for the following reaction is
= 542.11 kJ
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