Calculate the ΔH°f for Mg(NO3)2(s) from the following data:
8Mg(s) + Mg(NO3)2(s) → Mg3N2(s) + 6MgO(s)
ΔH° = -3280.88 kJ
Mg3N2(s) → 3Mg(s) + N2(g)
ΔH° = +461.08 kJ
2MgO(s) → 2Mg(s) + O2(g)
ΔH° = +1203.60 kJ
Solution Guide:
1. Write a balanced “target” reaction – standard enthalpy of formation for Mg(NO3)2(s)
Mg(s) + N2(g) + 3O2(g) → Mg(NO3)2(s)
2. Rearrange the given reactions to place the correct number of moles of each reactant and
product where they appear in the “target.” Wait to “fix” species that appear in more than one
reaction, if possible, to the end. They should work out.
Reverse the first to get Mg(NO3)2(s) on the right. The coefficient is right. Change the sign of ΔH°.
Mg3N2(s) + 6MgO(s) →8Mg(s) + Mg(NO3)2(s)
ΔH° = -3280.88 kJ x (-1) = +3280.88 kJ
Reverse the second to get N2(g)on the right. The coefficient is right. Change the sign of ΔH°.
3Mg(s) + N2(g) → Mg3N2(s)
ΔH° = +461.08 kJ x (-1) = -461.08 kJ
Reverse the third to get O2(g)on the right. Multiply the coefficients by 3 to get the “target”
value. Change the sign of ΔH° and multiply by 3.
6Mg(s) + 3O2(g) → 6MgO(s)
ΔH° = +1203.60 kJ x (-3) = -3610.80 kJ
3. Add the resulting set of equations by canceling reactants from one equation with products from
another (see above). If it doesn’t yield the “target” you did something wrong – I won’t give you a
problem that doesn’t work. The Mg(s) is a little tricky here, but it’s just arithmetic.
4. Add the enthalpies:
Mg(s) + N2(g) + 3O2(g) → Mg(NO3)2(s)
ΔH° = -791.00 kJ (2 decimal places)
Calculate the ΔH°f for dinitrogen pentoxide at 25 °C and 1 atm from the following data:
2H2(g) + O2(g) → 2H2O(ℓ)
ΔH = -571.6 kJ
N2O5(g) + H2O(ℓ) → 2HNO3(ℓ)
ΔH = -73.7 kJ
N2(g) + 3O2(g) +H2(g) → 2HNO3(ℓ)
ΔH = -348.2 kJ
Solution Guide:
1. Write a balanced “target” reaction – standard enthalpy of formation for N2O5(g).
N2(g) + 5/2 O2(g) → N2O5(g)
2. Rearrange the given reactions to place the correct number of moles of each reactant and
product where they appear in the “target.” Wait to “fix” species that appear in more than one
reaction, if possible, to the end. They should work out.
Reverse the first to get H2(g) on the right. The coefficient is unclear and must be revisited.
Change the sign of ΔH°.
2H2O(ℓ) → 2H2(g) + O2(g)
ΔH = -571.6 kJ x (-1) = +571.6 kJ
Reverse the second to get N2O5 (g)on the right. The coefficient is right. Change the sign of ΔH°.
2HNO3(ℓ) → N2O5(g) + H2O(ℓ)
ΔH = -73.7 kJ x (-1) = +73.7 kJ
The third equation needs no adjustment at this time (or possibly not at all)
N2(g) + 3O2(g) +H2(g) → 2HNO3(ℓ)
ΔH = -348.2 kJ
3.
Now we can see that some adjustments are needed somewhere. Let’s look at the “corrected” set
of equations one at a time:
2H2O(ℓ) → 2H2(g) + O2(g)
ΔH = -571.6 kJ x (-1) = +571.6 kJ
If we multiply this equation by ½, all our arithmetic problems are solved:
H2O(ℓ) → H2(g) + ½O2(g)
ΔH = -571.6 kJ x (½) = +285.8 kJ
4. Add the resulting set of equations by canceling reactants from one equation with products from
another:
H2O(ℓ) → H2(g) + ½O2(g)
ΔH = -571.6 kJ x (½) = +285.8 kJ
2HNO3(ℓ) → N2O5(g) + H2O(ℓ)
ΔH = -73.7 kJ x (-1) = +73.7 kJ
N2(g) + 3 5/2O2(g) +H2(g) → 2HNO3(ℓ)
ΔH = -348.2 kJ= -348.2 kJ
5. Add the enthalpies:
N2(g) + 5/2 O2(g) → N2O5(g)
ΔH° = 11.3 kJ (1 decimal place)
Calculate ΔH°rxn for the following reaction:
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)
Given:
ΔH0f
C6H12O6 (s)
CO2 (g)
H2O (l)
-1275.0 kJ/mol
-393.5 kJ/mol
-285.8 kJ/mol
ΔH°rxn = [6 mol(-393.5 kJ/mol) + 6 mol(-285.8 kJ/mol)] – [1 mol(-1275.0 kJ/mol) + 6 mol(0 kj/mol)]
ΔH°rxn = -2800.8 kJ