Chem 2 AP Homework #6-4: 46 The

Chem 2 AP Homework #6-4: Problems pg. 250-243 #46, 52, 54, 56, 62, 64, 110; Hess’s Law WS (optional)
46
The ΔH !f values of the two allotropes of oxygen, O2 and O3, are 0 and 142.2 kJ/mol,
respectively, at 25°C. Which is the more stable form at this temperature?
The standard enthalpy of formation of any element in its most stable form is zero. Therefore, since
!
ΔH f (O2 ) = 0, O2 is the more stable form of the element oxygen at this temperature.
52
The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily
assigning a value of zero to H+ ions; that is ΔH !f [H + (aq )] = 0 .
We use the ΔH !f values in Appendix 3 and Equation (6.18) of the text.
(a) For the following reaction, calculate ΔH !f for Cl– ions.
HCl(g) → H+(aq) + Cl–(aq)
∆H° = –74.9 kJ/mol
!
ΔH rxn
= ΔH !f (H + ) + ΔH !f (Cl − ) − ΔH !f (HCl )
−74.9 kJ/mol = 0 + ΔH !f (Cl–) − (1)(−92.3 kJ/mol)
!
−
ΔH f (Cl ) = − 167.2 kJ / mol
(b) Given that ΔH !f for OH– ions is –229.6 kJ/mol, calculate the enthalpy of neutralization
when a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as
KOH) at 25°C
The neutralization reaction is:
H+(aq) + OH–(aq) → H2O(l)
!
So, ΔH rxn
= ΔH !f [H 2O(l )] − [ΔH !f (H + ) + ΔH !f (OH − )]
ΔH !f [H 2O(l )] = − 285.8 kJ/mol (See Appendix 3 of the text.)
!
= (1)(−285.8 kJ/mol) − [(1)(0 kJ/mol) + (1)(−229.6 kJ/mol)] = −56.2 kJ/mol
ΔHrxn
54
Calculate the heats of combustion for the following reactions from the standard enthalpies of
formation listed in Appendix 3:
(a) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l)
ΔH° = [2ΔH !f (CO2 ) + 2ΔH !f (H 2O)] − [ΔH !f (C 2 H 4 ) + 3ΔH !f (O2 )] ]
ΔH° = [(2)(−393.5 kJ/mol) + (2)(−285.8 kJ/mol)] − [(1)(52.3 kJ/mol) + (3)(0)]
ΔH° = −1411 kJ/mol
(b) 2H2S(g) + 3O2(g) → 2H2O(l) + 2SO2(g)
ΔH° = [2ΔH !f (H 2O) + 2ΔH !f (SO2 )] − [2ΔH !f (H 2 S ) + 3ΔH !f (O2 )]
ΔH° = [(2)(−285.8 kJ/mol) + (2)(−296.1 kJ/mol)] − [(2)(−20.15 kJ/mol) + (3)(0)]
ΔH° = −1124 kJ/mol
2
56
HOMEWORK #6-4 ANSWER KEY
The standard enthalpy change for the following reaction is 436.4 kJ/mol:
→ H(g) + H(g)
H2(g) !!
Calculate the standard enthalpy of formation of atomic hydrogen (H).
!
ΔH rxn
= [ΔH f! (H ) + ΔH f! (H )] − ΔH f! (H 2 )
ΔH f! (H 2 ) = 0
!
ΔH rxn
= 436.4 kJ/mol = 2ΔH f! (H ) − (1)(0)
ΔH f! (H) =
62
436.4 kJ/mol
= 218.2 kJ / mol
2
From the following data, calculate the enthalpy change for the reaction
→ C2H6(g)
2C(graphite) + 3H2(g) !!
!
= −393.5 kJ/mol
ΔH rxn
!
ΔH rxn = −285.8 kJ/mol
!
= –3119.6 kJ/mol
ΔH rxn
→ CO2(g)
(a) C(graphite) + O2(g) !!
→ 3H2O(l)
(b) H2(g) + ½ O2(g) !!
→ 4CO2(g) + 6H2O(l)
(c) 2C2H6(g) + 7O2(g) !!
Looking at this reaction, we need two moles of graphite as a reactant. So, we multiply Equation (a) by
two to obtain:
→ 2CO2(g)
(d) 2C(graphite) + 2O2(g) !!
!
ΔH rxn = 2(−393.5 kJ/mol ) = − 787.0 kJ/mol
Next, we need three moles of H2 as a reactant. So, we multiply Equation (b) by three to obtain:
→ 3H2O(l)
(e) 3H2(g) + 32 O2(g) !!
!
ΔH rxn
= 3(−285.8 kJ/mol ) = − 857.4 kJ/mol
Last, we need one mole of C2H6 as a product. Equation (c) has two moles of C2H6 as a reactant, so we
need to reverse the equation and divide it by 2.
!
→ C2H6(g) + 72 O2(g) ΔH rxn
(f) 2CO2(g) + 3H2O(l) !!
= 12 (3119.6 kJ/mol ) = 1559.8 kJ/mol
Adding Equations (d), (e), and (f) together, we have:
Reaction
→ 2CO2(g)
(d) 2C(graphite) + 2O2(g) !!
3
→ 3H2O(l)
(e) 3H2(g) + 2 O2(g) !!
→ C2H6(g) + 72 O2(g)
(f) 2CO2(g) + 3H2O(l) !!
→ C2H6(g)
2C(graphite) + 3H2(g) !!
ΔH° (kJ/mol)
−787.0
−857.4
1559.8
ΔH° = −84.6 kJ/mol
HOMEWORK #6-4 ANSWER KEY
64
Calculate the standard enthalpy change for the reaction,
→ 2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s) !!
given that
→ Al2O3(s)
2Al(s) + 32 O2(g ) !!
ΔH° = −1601 kJ/mol
2Fe(s) +
3
2
→ Fe2O3(s)
O2(g) !!
ΔH° = –821 kJ/mol
The second and third equations can be rearranged and combined to give the first equation.
110
→ Al2O3(s)
2Al(s) + 32 O2(g ) !!
ΔH° = −1601 kJ/mol
→ 2Fe(s) + 32 O2(g)
Fe2O3(s) !!
ΔH° = 821 kJ/mol
→ 2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s) !!
ΔH° = −780 kJ/mol
At 25°C, the standard enthalpy of formation of HF(aq) is –320.1 kJ/mol; of OH– it is –229.6
kJ/mol; of F– it is –329.1 kJ/mol; and of H2O(l) it is –285.8 kJ/mol
(a) Calculate the standard enthalpy of neutralization of HF(aq):
HF(aq) + OH–(aq) → F–(aq) + H2O(l)
ΔH° = ΔH !f (F − ) + ΔH !f (H 2O) − [ΔH !f (HF ) + ΔH !f (OH − )]
ΔH° = [(1)(−329.1 kJ/mol) + (1)(−285.8 kJ/mol)] − [(1)(−320.1 kJ/mol) + (1)(−229.6 kJ/mol)
ΔH° = −65.2 kJ/mol
(b) Using the value of –56.2 kJ as the standard enthalpy change for the reaction
H+(aq) + OH–(aq) → H2O(l),
calculate the standard enthalpy change for the reaction
HF(aq) → H+(aq) + F–(aq)
We can add the equation given in part (a) to the reverse of that given in part (b) to end up with the
equation we are interested in.
→ F–(aq) + H2O(l)
HF(aq) + OH–(aq) !!
→ H+(aq) + OH–(aq)
H2O(l) !!
ΔH° = −65.2 kJ/mol
ΔH° = +56.2 kJ/mol
→ H+(aq) + F–(aq)
HF(aq) !!
ΔH° = −9.0 kJ/mol
3