CHM 3400 – Problem Set 3
Due date: NONE. (NOTE: The first exam is Wednesday, February 6 th. It will cover material from Chapters 1, 2,
and 3 of Atkins, plus handouts. The problems in this set cover material from Chapter 3.)
Do all of the following problems. Show your work.
1) The Haber process, which converts nitrogen and hydrogen into ammonia, is one of the most important reactions in
industrial chemistry. The reaction may be written as
N2(g) + 3 H2(g)  2 NH3(g)
(1.1)
a) How is reaction 1.1 related to the formation reaction for ammonia?
b) Find Hrxn for reaction 1.1 at T = 25. C.
c) Find Hrxn for reaction 1.1 at T = 450. C. Note that this is the temperature at which this reaction is
typically carried out industrially. You may assume in your calculation that Cp,m is independent of temperature for all
reactants and products.
2) Ethyl alcohol can be oxidized to ethanal (acetaldehyde) by the following reaction
2 CH3CH2OH(g) + O2(g)  2 CH3CHO(g) + 2 H2O(g)
(2.1)
a) Using the table of mean bond enthalpies in Atkins (Table 3.5, p 70) estimate the value for Hrxn for
reaction 2.1.
b) Using the thermochemical data in Appendix D1.1 and D1.2 of Atkins calculate the value for Hrxn for
reaction 2.1.
c) Which of the above two values for Hrxn is closest to the actual value? Explain.
Also do the following problems from Atkins:
3.19 The standard enthalpy of combustion of anthracene (C14H10(s)) is – 7163. kJ/mol. Calculate its standard
enthalpy of formation.
3.32 Calculate the standard enthalpy of formation of N2O5(g) from the following data
(1)
2 NO(g) + O2(g)  2 NO2(g)
H = - 114.1 kJ/mol
(2)
4 NO2(g) + O2(g)  2 N2O5(g)
H = - 110.2 kJ/mol
(3)
N2(g) + O2(g)  2 NO(g)
H = + 180.5 kJ/mol
Solutions.
1)
a) By definition the formation reaction is the reaction that produces one mole of a single product out of
elements in their standard (thermodynamically most stable) state. Reaction 1.1 produces two moles of ammonia out
of elements in their standard state, and so it is equal to two times the formation reaction for ammonia.
b) Based on the answer in part a, at T = 25.0 C
Hrxn = 2 Hf(NH3(g)) = 2 (- 46.11 kJ/mol) = - 92.22 kJ/mol
c) To do this problem we need an expression for Cp for the reaction. Since we are told we may assume
that the values for Cp,m for the reactants and products are all independent of temperature, then
Cp = 2 Cp,m(NH3(g)) – [ Cp,m(N2(g)) + 3 Cp,m(H2(g)) ]
= 2 (35.06) – [ (29.125) + 3 (28.824) ] = - 45.48 J/mol.K = - 45.48 x 10-3 kJ/mol.K
and so at T = 450. C, using the simplified form of Kirchoff’s law that applies when Cp is independent of
temperature
Hrxn (450. C) = Hrxn(25. C) + Cp (450. C – 25. C)
= - 92.22 kJ/mol + (- 45.48 x 10-3 kJ/mol) (425. K) = - 111.55 kJ/mol
Note that in the above calculation we have made use of the fact that the size of a degree is the same in the Centigrade
and Kelvin scale, so a temperature difference of 425. C is the same as a temperature difference of 425. K.
2)
a) We may estimate the value for Hrxn by find the bond enthalpies for the reactant and product molecules
bond enthalpies of reactants
bond enthalpies of products
10 (C-H) = 10 (412) =
2 (C-C) = 2 (348) =
2 (C-O) = 2 (360) =
2 (O-H) = 2 (463) =
1 (O=O) = 1 (497) =
8 (C-H) = 8 (412) =
2 (C-C) = 2 (348) =
2 (C=O) = 2 (743) =
4 (O-H) = 4 (463) =
4120
696
720
926
497
6957
3296
696
1486
1852
7330
So Hrxn  (6957 kJ/mol) – (7330 kJ/mol) = - 373 kJ/mol
b) We may find Hrxn in terms of the enthalpies of formation of the reactants and products
Hrxn = [ 2Hf(CH3CHO(g)) + 2 Hf(H2O(g)) ] – [ 2 Hf(C2H5OH(g)) + Hf(O2(g)) ]
= [ 2 (- 166.19) + 2 (- 241.82) ] – [ 2 (- 235.10) + 1 (0.0) ] = - 345.82 kJ/mol
Note that we really didn’t really need to include Hf(O2(g)) in the calculation, since it follows from the definition of
the formation reaction that the enthalpy of formation of an element in its standard state is exactly equal to zero.
c) The value calculated from the thermochemical data in part b should be nearly equal to the exact value, as
it is based on experimental data and Hess’ law. Since Hess’ law is exact, the only uncertainty will be due to
experimental uncertainty in the formation enthalpies, which is small. The value found in part a is based on average
bond enthalpies, which are approximate, and so the value for Hrxn calculated from the average bond enthalpies is
also expected to be only approximately correct.
3.19
The combustion reaction for anthracene is
C14H10(s) + 33/2 O2(g)  14 CO2(g) + 5 H2O()
Hrxn = Hcomb
From Hess’ law
Hcomb = 14 Hf(CO2(g)) + 5 Hf(H2O()) - Hf(C14H10(s))
So
Hf(C14H10(s)) = 14 Hf(CO2(g)) + 5 Hf(H2O()) - Hcomb
= 14 (- 393.51) + 5 (- 285.83) – (- 7163) = + 225. kJ/mol
3.32
The formation reaction for N2O5(g) is
N2(g) + 5/2 O2(g)  N2O5(g)
It can be obtained as follows
2 NO2(g) + ½ O2(g)  N2O5(g)
H = ½ (- 110.2 kJ/mol)
2 NO(g) + O2(g)  2 NO2(g)
H = (1) (- 114.1 kJ/mol)
N2(g) + O2(g)  2 NO(g)
___________________________
N2(g) + 5/2 O2(g)  N2O5(g)
H = (1) (+ 180.5 kJ/mol)
______________________
Hf(N2O5(g)) = + 11.3 kJ/mol