The production of ammonia (NH3) under standard conditions at 25 C is represented by the following thermochemical equation. N2(g)+3H2(g)=2NH3(g); heat of reaction=-91.8 kJ

How much heat in kJ is released when 1.671*10^4 grams of ammonia is produced?

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Given, Heat of formation =−46.0 KJ/mole

Since, the enthalpy of formation of ammonia is the enthalpy change for the formation of 1 mole of ammonia from its element.

2NH

3

(g)→2N

2

(g)+3H

2

(g)

△H

reaction

=2×△H

formation

=2×−46.0 KJ/mole

Heat of reaction=−92.0 KJ/mole

Explanation:

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the answer for sodium is 0.186

the answer for rubidium is 0.251

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