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Zaman University

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Chanpiset Um

on 13 June 2014

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Transcript of Zaman University

Ammonia (NH3) boils at -33 C ; at this temperature it has a density of 0.81g/cm3 . The enthalpy of formation of NH3(g) is -46.2kJ/mol, and the enthalpy of vaporization of NH3(l) is 23.2kJ/mol.
Calculate the enthalpy change when 1 L of liquid NH3 is burned in air to give N2(g) and H2O(g). How does this
compare with H for the complete combustion of 1 L of liquid
methanol, CH3OH(l)?
For CH3OH(l), the density at 25 C is 0.792g/cm3,and
Exercise 5.107
Zaman University

Chemistry Assignment
Instructor: Mustafa Dur

By: Chanpiset Um

The reaction for which we want to calculate the enthalpy change is:
Before we can calculate H for this reaction, we must calculate
H for NH3(l).
We know that enthalpy formation NH3(g) is -46.2kJ
And NH3(l) -> NH3(g)
Hvap= H(NH3(g))-
23.2kJ=-46.2kJ- H(NH3(l))
=> H(NH3(l)=-69.4kJ

Furthermore we know that
Since 1cm3 of NH3 weighs 0.81g, 1l=1000cm3 weighs 810g.
n(NH3)=m(NH3)/M(NH3) =(810/17)mol
Therefore combustion of 1L of NH3 which is 810/17 mol produces
Now we calculate enthalpy change of combustion of CH3OH(l)
The reaction is:
H=2 H(CO(g)) + 4 H(H2O(g)) - 2 H(CH3OH(g))-3 H(O2(g))
Then for the overall reaction, the enthalpy change is:
H=6 H(H2O(g)) + 2 H(N2(g)) - 4 H(NH3(l))-2 H(O2(g))

The result shows that combustion of 4 mol of NH3(l) produces 1173kJ of heat.
Now we need to find heat produced by 1L of NH3(l).
The result shows that:
combustion of 2 mol of CH3OH produces 1276kJ..
Now we need to find heat produced by 1L of CH3OH.
Since 1cm3 of CH3OH weighs 0.792g, 1L=1000cm3 weighs 792g.

Then heat produced by this amount of CH3OH is:
This results shows that methanol is a slight better fuel than ammonium liquid.
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