Calculate the enthalpy of formation (in kcal/mol) of gaseous HBr using following data:
Substance | NH₃(g) | HBr(g) | NH₄Br(s)
Heat of formation | -11 | X | -64
Heat of solution | -8.5 | -20.0 | +4.0 kcal
and, NH₃(aq) + HBr(aq) → NH₄Br(aq); ΔH = -12.3 kcal

Question

Calculate the enthalpy of formation (in kcal/mol) of gaseous HBr using following data:

Substance | NH₃(g) | HBr(g) | NH₄Br(s)

Heat of formation | -11 | X | -64

Heat of solution | -8.5 | -20.0 | +4.0 kcal

and, NH₃(aq) + HBr(aq) → NH₄Br(aq); ΔH = -12.3 kcal

TestHub · Accepted Answer

Go from NH 3 (g) + HBr(g) to NH 4 Br(s) via solution: dissolve NH 3 (-8.5), dissolve HBr (-20.0), react in solution (-12.3), crystallise NH 4 Br(aq) → NH 4 Br(s) (-4.0). Sum = -44.8 kcal. Directly, ΔH = -64 - (-11) - X = -53 - X. So -53 - X = -44.8 and X = -8.2 kcal/mol. This is option (b).

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