Bond Enthalpy Question - 4

QuestionA bond enthalpy calculation can be rearranged to find an unknown bond enthalpy when ΔH is known.H₂(g) + Br₂(g) → 2HBr(g) ΔH = -103 kJ mol^-1Bond enthalpies / kJ mol^-1: H-H 436, Br-Br 193. Calculate the H-Br bond enthalpy.Step 1: write the expression with x for the unknown.ΔH = Σ(bonds broken) - Σ(bonds made)
-103 = (436 + 193) - 2xStep 2: simplify.-103 = 629 - 2xStep 3: rearrange.2x = 629 + 103 = 732
x = +366 kJ mol^-1Check: put it back in. Broken 629 kJ, made 2 × 366 = 732 kJ, ΔH = 629 - 732 = -103 kJ mol^-1. Correct.Points to watchBromine must be gaseous, Br₂(g). At room temperature bromine is a liquid, and extra energy would be needed to vaporise it first, so the state symbol matters.Remember the factor of 2: two H-Br bonds are made, so the unknown appears as 2x. Forgetting it gives 732 kJ mol^-1, double the real value.A bond enthalpy is always positive. If your value of x is negative, a sign has gone wrong in the rearrangement.