Ammonia (NH3) boils at -33oC; at this temperature it has a density of 0.81 g/cm3. The enthalpy of formation of NH3(g) is -46.2kj/mol, and the enthalpy of vaporization of NH3(l) is 23.2 kJ/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 ΔH for the complete combustion of 1 L of liquid methanol, CH3OH(l)? For CH3OH(l), the density at 25oC is 0.792 g/cm3, and ΔHof equals -239 kJ/mol.
For NH3, 0.81 g/cc x 1000 cc = 810 grams.
810g/17 = ? mols.
2NH3 + 3/2 O2 ==> N2 + 3H2O
dHrxn = (3*dHf H2O + 0) - (2*dH NH3) = ? kJ/mol
You need to add in 23.2 to that to go from liquid NH3 to burned NH3.
Then total dHrxn (including vaporization) x mols NH3 = kJ.
Do the same thing for CH3OH but I see no number for correction from liquid to gas.
To calculate the enthalpy change when 1 L of liquid NH3 is burned in air to give N2(g) and H2O(g), we need to take into account the enthalpy change for the combustion reaction and the enthalpy change for the vaporization of NH3(l).
1. Calculate the number of moles of NH3:
Density = Mass/Volume
0.81 g/cm3 = Mass/1000 cm3
Mass = 0.81 g/cm3 * 1000 cm3 = 810 g
Molar mass of NH3 = 14 g/mol + 3(1 g/mol) = 17 g/mol
Number of moles = Mass/Molar mass = 810 g / 17 g/mol = 47.65 mol
2. Calculate the enthalpy change for the combustion reaction:
NH3(l) + air → N2(g) + H2O(g)
ΔHcombustion = ΔHof(N2(g)) + ΔHof(H2O(g)) - ΔHof(NH3(g))
ΔHof(N2(g)) = 0 kJ/mol (N2 is the standard state)
ΔHof(H2O(g)) = 0 kJ/mol (H2O is the standard state)
ΔHof(NH3(g)) = -46.2 kJ/mol (given)
ΔHcombustion= 0 kJ/mol + 0 kJ/mol - (-46.2 kJ/mol) = 46.2 kJ/mol
3. Calculate the enthalpy change for the vaporization of NH3:
ΔHvaporization = 23.2 kJ/mol (given)
4. Calculate the total enthalpy change for burning 1 L of liquid NH3:
Enthalpy change = ΔHcombustion + ΔHvaporization * Number of moles
Enthalpy change = 46.2 kJ/mol + 23.2 kJ/mol * 47.65 mol = 1104.78 kJ
Now, let's calculate the enthalpy change for the complete combustion of 1 L of liquid methanol, CH3OH(l):
1. Calculate the number of moles of CH3OH:
Density = Mass/Volume
0.792 g/cm3 = Mass/1000 cm3
Mass = 0.792 g/cm3 * 1000 cm3 = 792 g
Molar mass of CH3OH = 12 g/mol + 3(1 g/mol) + 16 g/mol + 1 g/mol = 32 g/mol
Number of moles = Mass/Molar mass = 792 g / 32 g/mol = 24.75 mol
2. Calculate the enthalpy change for the combustion reaction of methanol:
CH3OH(l) + O2(g) → CO2(g) + 2H2O(g)
ΔHcombustion = ΔHof(CO2(g)) + 2*ΔHof(H2O(g)) - ΔHof(CH3OH(l))
ΔHof(CO2(g)) = 0 kJ/mol (CO2 is the standard state)
ΔHof(H2O(g)) = 0 kJ/mol (H2O is the standard state)
ΔHof(CH3OH(l)) = -239 kJ/mol (given)
ΔHcombustion= 0 kJ/mol + 2*0 kJ/mol - (-239 kJ/mol) = 239 kJ/mol
3. Calculate the total enthalpy change for the complete combustion of 1 L of liquid methanol:
Enthalpy change = ΔHcombustion * Number of moles
Enthalpy change = 239 kJ/mol * 24.75 mol = 5912.25 kJ
Therefore, the enthalpy change for burning 1 L of liquid NH3 is 1104.78 kJ, whereas the enthalpy change for the complete combustion of 1 L of liquid methanol is 5912.25 kJ.
To calculate the enthalpy change when 1 L of liquid NH3 is burned in air, we need to consider the following steps:
Step 1: Calculate the mass of 1 L of liquid NH3.
Since the density of NH3 at -33oC is given as 0.81 g/cm3, we can convert the volume of 1 L to cm3 by multiplying by 1000.
Mass = Density x Volume = 0.81 g/cm3 x 1000 cm3 = 810 grams
Step 2: Calculate the number of moles of NH3.
We can use the molar mass of NH3 (17.03 g/mol) to convert the mass to moles.
Number of moles = Mass / Molar mass = 810 g / 17.03 g/mol = 47.6 mol
Step 3: Calculate the enthalpy change for the combustion of NH3.
The balanced chemical equation for the combustion of NH3 is:
4NH3 + 3O2 -> 2N2 + 6H2O
From the equation, we can see that the stoichiometric ratio between NH3 and N2 is 4:2. This means that for every 4 moles of NH3 burned, 2 moles of N2 are produced.
Therefore, the number of moles of N2 produced can be calculated as:
Number of moles of N2 = (2/4) x Number of moles of NH3 = (2/4) x 47.6 mol = 23.8 mol
The enthalpy change for the formation of N2(g) from its elements in their standard states is taken to be zero.
So, the enthalpy change for the combustion of 1 L of liquid NH3 to give N2(g) and H2O(g) can be calculated as follows:
Enthalpy change = ΔHof(N2(g)) + ΔHovap(NH3) - ΔHof(NH3(g)) - ΔHof(H2O(g))
Enthalpy change = 0 kJ/mol + (23.2 kJ/mol) - (-46.2 kJ/mol) - 0 kJ/mol
Enthalpy change = 23.2 kJ/mol + 46.2 kJ/mol
Enthalpy change = 69.4 kJ/mol
Now, let's compare this value with the enthalpy change for the complete combustion of 1 L of liquid methanol, CH3OH(l).
The balanced chemical equation for the combustion of methanol is:
CH3OH + 3/2 O2 -> CO2 + 2H2O
From the balanced equation, we can see that the stoichiometric ratio between methanol and CO2 is 1:1. This means that for every mole of methanol burned, one mole of CO2 is produced.
The enthalpy change for the combustion of methanol can be calculated as:
Enthalpy change = ΔHof(CO2) - ΔHof(CH3OH)
Enthalpy change = 0 kJ/mol - (-239 kJ/mol)
Enthalpy change = 239 kJ/mol
Therefore, the enthalpy change for the complete combustion of 1 L of liquid methanol is 239 kJ/mol, which is higher than the enthalpy change for the combustion of 1 L of liquid NH3 (69.4 kJ/mol).