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The energy change ∆H associated with the reaction NBr3(g) + 3 H2O(g) → 3 HOBr(g) + NH3(g) is +81 kJ. NBr3 is arranged with all of the bromines attached to the nitrogen and HOBr is arranged as H O Br. The O H bond is 459 kJ The N H bond is 386 kJ The O H bond is 459 kJ The O Br bond is 201 kJ The strength of the N Br bond is

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IthaloAbreu

Answer:

155 kJ

Explanation:

The energy change will be the energy of the reactants less the energy of the products. And the energy of each compound is the sum of the energy of their bonds. Let's call y the N-Br strength.

3HOBR = 3x(459 + 201) = 1980 kJ

NH₃ = 3x386 = 1158 kJ

3H₂O = 3x(2x459) = 2754 kJ

NBr₃ = 3y

3y + 2754 - 1158 - 1980 = 81

3y -384 = 81

3y = 465

y = 155 kJ

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