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?
1.) -81 kJ
2.) +162 kJ
3.) +243 kJ
4.) -365 kJ
5.) -243 kJ
6.) +365 kJ
7.) +81 kJ
8.) -162 kJ

Answer 1

Expected energy change is -162kJ

Step-by-step explanation:

For the reaction:

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

For the reverse reaction, ΔH changes sign, thus:

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

If 2 moles of NH₃ react, the ΔH must be multiplied twice:

6HOBr(g) + 2NH₃(g) → 2NBr₃(g) + 6H₂O(g) ΔH = -162kJ

As you have 9 moles of HOBr, the limitng reactant is NH₃. Thus, expected energy change is -162kJ

I hope it helps!