Question

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?

Answer 1

162 kJ

Step-by-step 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.