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:

If we turn it around, we have:

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:

-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.

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