### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

You can change this under Settings & Account at any time.

# Zaman University

No description
by

## Chanpiset Um

on 13 June 2014

Report abuse

#### 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
Hf=-239kJ/mol.
Exercise 5.107
Zaman University

Chemistry Assignment
Instructor: Mustafa Dur

By: Chanpiset Um

Solution
The reaction for which we want to calculate the enthalpy change is:
4NH3(g)+3O2(g)->2N2(g)+6H2O(g)
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.=23.3kJ
Thus,
Hvap= H(NH3(g))-
H(NH3(l)
23.2kJ=-46.2kJ- H(NH3(l))
=> H(NH3(l)=-69.4kJ

Furthermore we know that
H(H2O(g)=-241.82kJ
H(O2(g))=0
H(N2(g))=0
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
(1173x(810/17))/4=13972kJ
Now we calculate enthalpy change of combustion of CH3OH(l)
The reaction is:
2CH3OH(l)+3O2(g)
->2CO2(g)+4H2O(g)
H=2 H(CO(g)) + 4 H(H2O(g)) - 2 H(CH3OH(g))-3 H(O2(g))
=2(-393.5)+4(-241.82)-2(-239)-3(0)
=-1276kJ
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))
=5(-241.83)+2(0)-4(-69.4)-3(0)
=-1173kJ

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.

n(CH3OH)=m/M=792/32=24.75mol
Then heat produced by this amount of CH3OH is:
(1276x24.75)/2=1790.5kJ
This results shows that methanol is a slight better fuel than ammonium liquid.
Full transcript