Question

The enthalpy of reaction for the decomposition of ammonium nitrate, NH4NO3 is -77.4 kJ mol^-1. Calculate the heat evolved when 78.8 g of water is formed from this reaction. Given: NH4NO3 (S)—> N2O+ 2H2O (g)

Answer 1

Answer:

- 169.42 kJ.

Explanation:

• We have the enthalpy of reaction for the decomposition of ammonium nitrate, NH₄NO₃, is - 77.4 kJ/mol.

• We need to calculate the no. of moles of NH₄NO₃ that will produce 78.8 g of H₂O.

• Firstly, we need to calculate the no. of moles of 78.8 g of H₂O produced using the relation:

n = mass/molar mass.

∴ n of H₂O = mass/molar mass = (78.8 g)/(18.0 g/mol) = 4.377 mol.

• From the balanced equation of the decomposition of NH₄NO₃:

NH₄NO₃ → N₂O + 2H₂O,

It is clear that every 1.0 mole of NH₄NO₃ decomposes to 1.0 mole of N₂O and 2.0 moles of H₂O.

Using cross multiplication:

1.0 mole of NH₄NO₃ decomposes into → 2.0 moles of H₂O.

??? mole of NH₄NO₃ decomposes into → 4.377 moles of H₂O.

∴ The no. of moles of NH₄NO₃ that will produce 78.8 g of H₂O = (1.0 mol)(4.377 mol)/(2.0 mol) = 2.188 mol.

• To get the heat evolved when 2.188 mol of NH₄NO₃ decomposed:

1.0 mole of NH₄NO₃ gives → - 77.4 kJ/mol.

∴ The heat evolved when 78.8 g of water is formed from this reaction = (no. of moles of NH₄NO₃)(heat evolved from decomposition of 1.0 mole of NH₄NO₃) = (2.188 mol)(- 77.4 kJ/mol) = - 169.42 kJ.