Question

Ammonia (NH3) boils at -33∘C; at this temperature it has a density of 0.81 g/cm3. The enthalpy of formation of NH3(g) is -46.2 kJ/mol, and the enthalpy of vaporization of NH3(l) is 23.2 kJ/mol Calculate the enthalpy change when 5 L of liquid NH3 is burned in air to give N2(g) and H2O(g).

Answer 1

Answer:

-87098.82kJ

Explanation:

Hello,

This process could be attained into two steps due to the specified conditions at assuming that it is at -33°C at the beginning:

Δ$H=$Δ$vapH+$Δ$rH$

Thus:

Δ$vapH=5L*0.81\frac{kg}{L}*\frac{1000g}{1kg}*\frac{1molNH_3}{17gNH_3}*23.2\frac{kJ}{molNH_3}=5527.06kJ$

Once the ammonia is gaseous, the burning chemical reaction is:

$NH_3+\frac{3}{4} O_2-->\frac{1}{2}N_2+\frac{3}{2}H_2O$

So the enthalpy of reaction for the given amount of ammonia is:

Δ$rH=5L*0.81\frac{kg}{L}*\frac{1000g}{1kg}*\frac{1molNH_3}{17gNH_3}*(\frac{3}{2}*-290\frac{kJ}{mol}-(-46.2\frac{kJ}{mol}) ) =-92625.88kJ$

Finally, the total enthalpy is:

Δ$H=5527.06kJ-92625.88kJ=-87098.82kJ$

Best regards.