How many kilojoules are associated with the formation of 2 moles of HBr (g)

HBr is formed from hydrogen gas and bromine gas as shown in the reaction below;

H2 (g) + Br2 (g) = 2HBr (g)

The reaction involves bond breaking and bond formation

Decomposition of 1 mole of HBr requires 36.4kj/mol,

But, energy used in decomposition = energy used in formation

Thus, formation of a mole of HBr is - 36.4 kj/mol

Therefore, formation of two moles of HBr would require

(-36.4 * 2) = 72.8

≈ - 73 kj/mol

Hence, the enthalpy of formation of 2 moles of HBr (ΔH) is - 73 kj/mol