Question

How much heat energy (in kJ) will be absorbed or released if 8.33 grams of ammonia is produced?

Answer 1

Answer:

− 68,500 kJ

Explanation:

The problem provides you with the thermochemical equation that describes the synthesis of ammonia from nitrogen gas and hydrogen gas.

N2(g)+3H2(g)→2NH3(g)ΔH∘rxn=−92.6 kJ mol−1

Notice that the standard enthalpy change of reaction,

ΔH∘rxn, is equal to −92.6 kJ mol−1. This tells you that when 1 mole of ammonia is produced by the reaction, 92.6 kJ of heat are being given off.

Keep in mind that the minus sign attached to the standard enthalpy change of reaction signifies heat lost.

ΔHrxn= 92.6 kJ.mol−1 ⇒92.6 kJ.lost.per mole.of NH3 produced

Your first step will be to convert the number of grams of ammonia to moles by using the compound's molar mass 1.26⋅104g⋅ 1 mole NH^3 17.031g= 739.83 moles NH^3

So, if 92.6 kJ  of heat are being given off when 1mole is produced, it follows that when 739.83 moles are being produced, the reaction will give off 739.83 moles NH^3⋅92.6 kJ 1mole NH3= 68,500 kJ

−−−−−−−−

The answer is rounded to three sig figs.

Keep in mind that when 68,500 kJ  of heat are being given off, the standard enthalpy change of reaction must carry a minus sign!

ΔH∘rxn for 739.83 moles NH3= −68,500 kJ