Question

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

Answer 1

(-36.4 ×2) = 72.8

≈ -73 kj/mol

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

Answer 2

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