Determine the amount of heat (in kJ) associated with the production of 5.71 × 104 g of ammonia according to the following equation. N2(g) + 3H2(g) 2NH3ΔH°rxn = −92.6 kJ Assume that the reaction takes place under standard-state conditions at 25°C.

In this case, for the given reaction, we are given the standard enthalpy of reaction per mole of ammonia that is -92.6 kJ, it means, that forming one mole of ammonia will release 92.6 kJ of energy. In such a way, for the formation of 5.71x10⁴ g of ammonia, the following amount of heat will be released:

[tex]Q=5.71x10^4gNH_3*\frac{1molNH_3}{17gNH_3}*-92.6\frac{kJ}{molNH_3}\\ \\Q=-3.11x10^5kJ[/tex]

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Eduard22sly Eduard22sly · Answer 2 · 2021-11-18T11:03:52+01:00

The amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

We'll begin by calculating the number of mole in 5.71×10⁴ g of NH₃

Mass of NH₃ = 5.71×10⁴ g

Molar mass of NH₃ = 14 + (3×1) = 17 g/mol

Mole of NH₃ =?

Mole = mass / molar mass

Mole of NH₃ = 5.71×10⁴ / 17

Mole of NH₃ = 3358.82 moles

Finally, we shall determine the heat required to produce 3358.82 moles (i.e 5.71×10⁴ g) of NH₃. This can be obtained as follow:

N₂(g) + 3H₂(g) —> 2NH₃(g) ΔH°rxn = −92.6 kJ

Since reaction took place at standard conditions, it means:

1 moles of NH₃ required −92.6 kJ

Therefore,

3358.82 moles of NH₃ will require = 3358.82 × –92.6 = –311026.732 KJ

Thus, the amount of the heat associated with the production of 5.71×10⁴ g of ammonia, NH₃ is –311026.732 KJ

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