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Using standard heats of formation, calculate the standard enthalpy change for the following reaction.

2BrF3(g) -----> Br2(g) + 3F2(g)

Sagot :

okpalawalter8

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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