Bromine and hydrogen react together to form hydrogen bromide - Scottish Highers Chemistry - Question 18 - 2017

To calculate the overall enthalpy change for the reaction, we first need to determine the total energy required to break the bonds in the reactants and the total energy released by forming the bonds in the products.

1. Calculate the energy for bonds broken:

   • 1 bond of H–H: +436 kJ/mol
   • 1 bond of Br–Br: +194 kJ/mol

   Total energy for bonds broken:

   $E_{broken} = 436 + 194 = 630 ext{ kJ/mol}$

2. Calculate the energy for bonds made:

   • 2 bonds of H–Br: -2 × 366 kJ/mol = -732 kJ/mol

   Total energy for bonds made:

   $E_{made} = -732 ext{ kJ/mol}$

3. Calculate the enthalpy change (ΔH):

   $ΔH = E_{broken} - E_{made}$
   $ΔH = 630 - 732 = -102 ext{ kJ/mol}$

Thus, the enthalpy change for the reaction is -102 kJ/mol.