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2 October, 17:57

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?

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  1. 2 October, 18:42
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    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.
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