Answer
$ \Delta H^{\circ}_rxn = - 290 kJ/mol = -2.90\times10^{2}kJ/mol$
Work Step by Step
Equation (1) $Br_{2(l)} + F_{2(g)} →2BrF{(g)}$
$ \Delta H^{\circ}_rxn(1) = - 188 kJ/mol$
Equation (2) $Br_{2(g)} + 3 F_{2(g)} →2 BrF_{3(g)}$
$ \Delta H^{\circ}(rxn2) = - 768 kJ (/mol$
Divide Equation 1 by 2 and reverse to get Equation (3)
Equation (3) $ BrF{(g)} → (1/2) Br_{2(l)} +(1/2) F_{2(g)}$
$ \Delta H^{\circ}_rxn(3) = - (- 94 kJ/mol) = + 94 kJ/mol$
Divide Equation 2 by 2 to get Equation (4)
Equation (4) $(1/2)Br_{2(g)} + (3/2) F_{2(g)} → BrF_{3(g)}$
$ \Delta H^{\circ}_rxn(4) = - 384 kJ/mol$
The required equation,
$ BrF{(g)} + F_{2(g)} → BrF_{3(g)} $ is formed by adding Equation (3) and Equation (4).
$ \Delta H^{\circ}_rxn = \Delta H^{\circ}rxn(3) +\Delta H^{\circ}rxn(4)
= (94 + - 384 ) kJ/mol = - 290 kJ/mol = -2.90\times10^{2}kJ/mol$
