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tino4ka555 [31]
4 years ago
11

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 k

J/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).
Chemistry
1 answer:
Minchanka [31]4 years ago
7 0

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:

ΔH=ΔvapH+ΔrH

Thus:

ΔvapH=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:

ΔrH=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:

ΔH=5527.06kJ-92625.88kJ=-87098.82kJ

Best regards.

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