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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 :

sebassandin

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[tex]H=[/tex]Δ[tex]vapH+[/tex]Δ[tex]rH[/tex]

Thus:

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

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

[tex]NH_3+\frac{3}{4} O_2-->\frac{1}{2}N_2+\frac{3}{2}H_2O[/tex]

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

Δ[tex]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[/tex]

Finally, the total enthalpy is:

Δ[tex]H=5527.06kJ-92625.88kJ=-87098.82kJ[/tex]

Best regards.

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