CHM 3400 – Problem Set 3

Due date: NONE. (NOTE: The first exam is Wednesday, February 6th. 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 DH°rxn for reaction 1.1 at T = 25. °C.

c) Find DH°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 DH°rxn for reaction 2.1.

b) Using the thermochemical data in Appendix D1.1 and D1.2 of Atkins calculate the value for DH°rxn for reaction 2.1.

c) Which of the above two values for DH°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) DH° = - 114.1 kJ/mol

(2) 4 NO2(g) + O2(g) ® 2 N2O5(g) DH° = - 110.2 kJ/mol

(3) N2(g) + O2(g) ® 2 NO(g) DH° = + 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

DH°rxn = 2 DH°f(NH3(g)) = 2 (- 46.11 kJ/mol) = - 92.22 kJ/mol

c) To do this problem we need an expression for DCp 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

DCp = 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 DCp is independent of temperature

DH°rxn (450. °C) = DH°rxn(25. °C) + DCp (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 DH°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) = 4120 8 (C-H) = 8 (412) = 3296

2 (C-C) = 2 (348) = 696 2 (C-C) = 2 (348) = 696

2 (C-O) = 2 (360) = 720 2 (C=O) = 2 (743) = 1486

2 (O-H) = 2 (463) = 926 4 (O-H) = 4 (463) = 1852

1 (O=O) = 1 (497) = 497 7330 6957

So DH°rxn @ (6957 kJ/mol) – (7330 kJ/mol) = - 373 kJ/mol

b) We may find DH°rxn in terms of the enthalpies of formation of the reactants and products

DH°rxn = [ 2DH°f(CH3CHO(g)) + 2 DH°f(H2O(g)) ] – [ 2 DH°f(C2H5OH(g)) + DH°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 DH°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 DH°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() DH°rxn = DH°comb

From Hess’ law

DH°comb = 14 DH°f(CO2(g)) + 5 DH°f(H2O()) - DH°f(C14H10(s))

So DH°f(C14H10(s)) = 14 DH°f(CO2(g)) + 5 DH°f(H2O()) - DH°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) DH° = ½ (- 110.2 kJ/mol)

2 NO(g) + O2(g) ® 2 NO2(g) DH° = (1) (- 114.1 kJ/mol)

N2(g) + O2(g) ® 2 NO(g) DH° = (1) (+ 180.5 kJ/mol)

______

N2(g) + 5/2 O2(g) ® N2O5(g) DH°f(N2O5(g)) = + 11.3 kJ/mol