Answer
$\Delta H^{\circ}rxn(5) = - 350.657 kJ/mol = - 3.51/times 10^{2}kJ/mol$
Work Step by Step
Equation (1) $H_{2(g)} →2H{(g)} $
$ \Delta H^{\circ}_rxn(1) = 436.4 kJ/mol$
Equation (2) $Br_{2(g)} →2Br_{(g)}$
$ \Delta H^{\circ}_rxn(2) = 192.5 kJ/mol$
Equation (3) $H_{2(g}) + Br_{2(g)} → 2HBr_{(g)}$
$ \Delta H^{\circ}_rxn(3) = - 72.4 kJ/mol$
Divide equation (1) by 2 and reverse to get Equation (4)
Equation (4) $ H{(g)} → (1/2) H_{2(g)}$
$ \Delta H^{\circ}_rxn(4) = - 218.2 kJ/mol$
Divide equation (2) by 2 and reverse to get Equation (5)
Equation (5) $ Br{(g)} → (1/2) Br_{2(g)}$
$ \Delta H^{\circ}_rxn(5) = - 96.25 kJ/mol$
Divide equation (3) by 2 to get Equation (6)
Equation (6) $(1/2)H_{2(g}) +(1/2) Br_{2(g)} → HBr_{(g)}$
$ \Delta H^{\circ}_rxn(5) = - 36.2 kJ/mol$
The required equation,
$ H{(g)} + Br{(g)} → HBr_{(g)} $ is formed by adding Equation (4),Equation (5), and Equation (6).
$ \Delta H^{\circ}_r = \Delta H^{\circ}rxn(2) + \Delta H^{\circ}rxn(4) + \Delta H^{\circ}rxn(5) = (- 218.2+ - 96.25 +- 36.2) kJ/mol = - 350.657 kJ/mol = - 3.51/times 10^{2}kJ/mol$
