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Kruka [31]
3 years ago
11

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 exp

ected energy change for the reverse reaction of nine moles of HOBr and two moles of NH3?
Chemistry
1 answer:
xeze [42]3 years ago
6 0

Answer:

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 <u>which one will be the </u><em><u>limiting reactant</u></em> 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 <em>reactant in excess</em>, thus NH₃ is our limiting reactant.

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

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Explanation:


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2) Colligative properties are those physical properties of solutions that depends on the number of solute particles dissolved into the solution.


3) The relation between the number of solute particles and the depresson of the freezing point is proportional: the greater the number of solute particles the greater the freezing point depression.


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Note:

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