Question

Ammonium nitrate (NH4NO3(s) decomposes into nitrogen, oxygen and water. If 10 g of ammonium nitrate decompose entirely at 350 *C with ΔrxnHo = -289. kJ

- How much heat is liberated?

- How much work would be done by the system?

- What will be the pressure of the system if, after completion of the reaction, the volume is 1.0 L and the temperature is 120°C?

Answer 1

NH₄NO₃       N₂   + 1/2O₂ + 2H₂O          H^o_rxn = -289KJ

Here H^o_rxn is the amount of heat released when one mole of NH₄NO₃ decomposes

Here 10 g of NH₄NO₃ decomposes

10 g / 80.043 g/mol = 0.1249 mol

So heat liberated is -289 x 0.1249 = 36.105 KJ

Work

W = -PV = -pV

= -nRT

=-0.1249 mol x 8.3145 J K^-1mol^-1 x 623.15 K

Work = -0.647 KJ

Pressure of the system:

Assuming ideal gas situation we have PV = nRT

P = nRT/V

NH₄NO₃       N₂   + 1/2O₂ + 2H₂O

Since the reaction is complete 1 mol of Ammonium Nitrate gives 1 mol of N2 and 1/2 mol of O2 and 2 mol of water.

So, 0.1249 mol of Ammonium Nitrate gives 0.1249 mol of N2 and 0.1249/2 mol of O2

and 2x0.1249 mol of water

Total number of moles after decomposition is complete is

= 0.1249 + 0.1249/2 + 2 x 0.1249

=0.43715mol

P =nRT/V

P = 0.43715 mol x 8.3145 x 10^-2 bar K^-1mol^-1 x 393.15 K / 1.0 dm³

P = 14.3 bar