The energy change, ∆H, associated with the following reaction is + 81 kJ. NBr3 (g) + 3 H2O (g) → 3 HOBr (g) + NH3 (g) What is the expected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?

162 kJ

Explanation:

The reaction given by the problem is:

NBr₃ (g) + 3 H₂O (g) → 3 HOBr (g) + NH₃ (g) ∆H = + 81 kJ

If we turn it around, we have:

3 HOBr (g) + NH₃ (g) → NBr₃ (g) + 3 H₂O (g) ∆H = - 81 kJ

If we think now of HOBr and NH₃ as our reactants, then now we need to find out which one will be the limiting reactant when we have 9 moles of HOBr and 2 moles of NH₃:

When we have 1 mol NH₃, we need 3 mol HOBr. So when we have 2 moles NH₃, we need 6 moles HOBr. We have more than 6 moles HOBr so that's our reactant in excess, thus NH₃ is our limiting reactant.

-81 kJ is our energy change when there's one mol of NH₃ reacting, so we multiply that value by two when there's two moles of NH₃ reacting. The answer is 81*2 = 162 kJ.