Ch. 5

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enthalpy, or just bond energy. We would write it as BE(H–H) = 436.0 kJ/mol. We can use the bond-energies to calculate. (approximate) enthalpies of formation ...
Dr. Pedro Julio Villegas Aguilar [email protected]

OBJECTIVES • To become familiar with problems about thermo-chemistry (Hees’s Law and Bond Energies). TEXTBOOK Brown, Lemay & Bursten, Chemistry: The Central Science, 10th Ed. (Chapter 5)

QUESTION 1: What is Hess’s Law?

HESS’S

LAW

• If a reaction is carried out in a series of steps, ∆H for the reaction is the sum of ∆H for each of the steps. • The total change in enthalpy is independent of the number of steps. • Total ∆H is also independent of the nature of the path. • Hess’s Law provides a useful means of calculating energy changes that are difficult to measure.

HESS’S

LAW

Finding DH for a reaction when DHf of all reactants and products is known. DH = [Sum DHf products] – [Sum DHf reactants]

Remember that DHf of all elements is zero. Also, watch for the very frequent mistake of doing reactants – products, rather than products – reactants.

QUESTION 2: Give a brief explanation about bond energies.

QUESTION 2: Give a brief explanation about bond energies. • Energy is supplied to break the chemical bonds and energy is always released when the chemical bonds are formed. Therefore, the change in enthalpy of any chemical reactions in terms of bond energies is given by:

DH0 = ∑ Bond energies of reactants – ∑Bond energies of products =Total input energy–Total output energy

BOND ENERGIES When bonds are formed energy is released. Electrons are in a more stable arrangement when in a bond (molecular orbital) than when they are unpaired (non-bonded) in atomic orbital. H2(g) → 2H (g,atom) DH° = 436.0 kJ/mol This is a bond-dissociation enthalpy or, bond enthalpy, or just bond energy. We would write it as BE(H–H) = 436.0 kJ/mol We can use the bond-energies to calculate (approximate) enthalpies of formation for any compound. Take, for example H(g,atom). DHf° = ½ BE(H–H). ½H2(g) → H(g,atom) DHf° = 218.0 kJ/mol

BOND ENERGIES Consider the bonds in methane CH4. There are four C-H bonds. CH4(g) → C(g,atom) + 4 H(g,atom) DH° = 4 BE(C–H) DH°

= DHf°(Cg,atom) + 4DHf°(Hg,atom) - DHf°(CH4g) = 716.7kJ/mol+4(218.0 kJ/mol)-(-74.5 kJ/mol) = 1663 kJ/molBE(C–H)

= DH°/4 = 1663 kJ/mol/4 = 415.8 kJ/mol

BOND ENERGIES Now, let's consider the bonds in C2H6. There is one C-C bond and there are 6 C-H bonds. C2H6(g) → 2 C(g,atom) + 6 H(g,atom) DH° = 2DHf°(Cg,atom) + 6DHf°(Hg,atom) - DHf°(C2H6g) =2(716.7kJ/mol)+6(218.0kJ/mol)-(-84.7 kJ/mol) = 2826.1 kJ/mol We assume BE(C–H) = 415.8 kJ/mol (same as for CH4). DH° = BE(C–C) + 6BE(C–H) = DH° - 6 BE(C–H)

= 2826.1 kJ/mol - 6*415.8 kJ/mol = 331.3 kJ/mol

PROBLEM 1: • Given the following bond dissociation energies (H-C is 413 kJ/mol; H-H is 436 kJ/mol; C=C is 614 kJ/mol; C-C is 348 kJ/mol), determine DH for the reaction: H2C=CH2(g) + H2(g) → H3C-CH3(g)

PROBLEM 1: Given the following bond dissociation energies (H-C is 413 kJ/mol; H-H is 436 kJ/mol; C=C is 614 kJ/mol; C-C is 348 kJ/mol), determine DH for the reaction: H2C=CH2(g) + H2(g) → H3C-CH3(g)

•a 1

PROBLEM 2: Calculate DHf of CCl4(l) given the following data: CCl4(l) → CCl4(g) DH° = +31 kJ/mol C(s) → C(g) DH° = +715 kJ/mol Bond enthalpy (Cl-Cl) = +242 kJ/mol Bond enthalpy (C-Cl) = +338 kJ/mol

PROBLEM 2:Calculate DHf of CCl4(l) given the following data: CCl4(l) → CCl4(g) DH° = +31 kJ/mol C(s) → C(g) DH° = +715 kJ/mol Bond enthalpy (Cl-Cl) = +242 kJ/mol Bond enthalpy (C-Cl) = +338 kJ/mol

Analize situation step by step: C(s)

→ C(g)

2Cl2(g) → 4Cl(g)

DH = +715 kJ/mol DH = 2(+242 kJ/mol)

C(g) + 4Cl(g) → CCl4(g) DH = 4(-338 kJ/mol) CCl4(g) → CCl4(l)

C(s) + 2Cl2(g) → CCl4(l)

DH = -31 kJ/mol

DHf = -184kJ/mol

PROBLEM 3: Calculate the bond enthalpy of the C-N bond in CH3NH2(g) given the following data: C(s) → C(g) DH° = +715 kJ/mol Bond enthalpy: (C-H) = +412 kJ/mol Bond enthalpy: (N-H) = +388 kJ/mol Bond enthalpy: (H-H) = +436 kJ/mol Bond enthalpy: (N≡N) = +944 kJ/mol DHf CH3NH2(g) = –23 kJ/mol

PROBLEM 3:Calculate the bond enthalpy of the C-N bond in CH3NH2(g) given the following data: C(s) → C(g) DH° = +715 kJ/mol Bond enthalpy: (C-H) = +412 kJ/mol; (N-H) = +388 kJ/mol; (H-H) = +436 kJ/mol; (N≡N) = +944 kJ/mol and DHf CH3NH2(g) = –23 kJ/mol

Let do it step by step: C(s)+2½H2(g)+½N2(g) → CH3NH2(g) DHf = -23 kJ/mol DHconsumed, is related with: Breaking bonds: H-H and N≡N And also: C(s) → C(g) DH° = +715 kJ/mol DHconsumed = 715 +2½(436) + ½(944) = 2277 kJ DHreleased, is related with: Formed bonds: C-H, N-H and C-N And also: compound formation DHf = -23 kJ/mol DHreleased = 3(412) + 2(388) + DHC-N - 23= 1989 kJ+DHC-N DHreleased = DHconsumed 2282 kJ = 1989 kJ + DHC-N DHC-N = 288 kJ/mol

PROBLEM 4: Determine the enthalpy change for the reaction: 2CO + O=O → 2 O=C=O given the following bond energies: BE(CO) = 1074 kJ; BE(O=O) = 499 kJ; BE(C=O) = 802 kJ.

PROBLEM 4: Determine the enthalpy change for the reaction: 2CO + O=O → 2 O=C=O given the following bond energies: BE(CO) = 1074 kJ BE(O=O) = 499 kJ BE(C=O) = 802 kJ.

Lets evaluate what happened here:

Breaking bonds: 2CO and O=O Formed bonds: 4 C=O ΔH = [2(1074) + 499] – [4(802)] ΔH = -561 kJ

PROBLEM 5: Find DH for the following reaction using the data below. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) DHf(C3H8(g)) = –104 kJ/mol DHf(CO2(g)) = –394 kJ/mol DHf(H2O(l)) = –286 kJ/mol

PROBLEM 5: Find DH for the following reaction using the data below. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) DHf: C3H8(g) = –104; CO2(g) = –394; H2O(l) = –286 kJ/mol

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) DH= [Sum DHf products]–[Sum DHf reactants] = [ 3(-394) + 4(-286) ] – [–104] = –2 222 kJ/mol

PROBLEM 6: • Determining DH for the reaction: Given:

PROBLEM 6: Determining DH for the reaction: Given:

A way to look at the method of combining reactions would be as follows:

C(s) + ½O2(g) → CO(g) DH = (-393.5 kJ) + (283.0 kJ) = -110.5 kJ

PROBLEM 7: Calculate the heat change for the formation of 1 mol of strontium carbonate from it's elements. Given the following experimental information:

PROBLEM 7: Calculate the heat change for the formation of 1 mol of strontium carbonate from it's elements. Given the following experimental information:

1.

2.

3. 4.

If we are going to add these equations to get our desired equation, we must first have a desired equation. You must ask yourself: what are we looking for? The answer is the heat of formation of SrCO3. So, we must write an equation which represents the heat of formation of SrCO3. What is heat of formation? It is the heat required to make 1 mole of a compound from its elements in their standard states. We are trying to make SrCO3. It is formed from Sr, C and O2. Therefore, the balanced equation is:

5. Now, looking at the experimental data, can we add those equations together to get this equation? 6. The answer is yes!

PROBLEM 7: Calculate the heat change for the formation of 1 mol of strontium carbonate from it's elements. Given the following experimental information:

PROBLEM 8: Calculate the enthalpy for the following reaction: N2(g) + 2O2(g) → 2NO2(g) DH° = ??? kJ Using the following two equations: N2(g) + O2(g) → 2NO(g)

DH° = +180 kJ

2NO2(g) → 2NO(g) + O2(g) DH° = +112 kJ

PROBLEM 8: Calculate the enthalpy for the following reaction: N2(g) + 2O2(g) → 2NO2(g) DH° = ??? kJ Using the following two equations: N2(g) + O2(g) → 2NO(g) DH° = +180 kJ 2NO2(g) → 2NO(g) + O2(g) DH° = +112 kJ

Answer: N2(g) + O2(g) → 2NO(g) DH° = +180 kJ 2NO(g) + O2(g) → 2NO2(g) DH° = -112 kJ Notice that I have also changed the sign on the enthalpy from positive to negative. Next, we add the two equations together and eliminate identical items. We also add the two enthalpies together. N2(g) + 2O2(g) → 2NO2(g) DH° = +68 kJ

PROBLEM 9: Calculate DH° for this reaction:

2N2(g) + 5O2(g) → 2N2O5(g) Using the following three equations: H2(g) + (½) O2(g) → H2O(l) N2O5(g) + H2O(l) → 2HNO3(l)

DH°=-285.8 kJ DH°=-76.6 kJ

(½) N2(g)+(3/2)O2(g)+(½)H2(g) → HNO3(l)

DH°=-174.1 kJ

PROBLEM 9: Calculate DH° for this reaction: 2N2(g) + 5O2(g) → 2N2O5(g) Using the following three equations: H2(g) + (½) O2(g) → H2O(l)

DH°=-285.8 kJ

N2O5(g) + H2O(l) → 2HNO3(l)

DH°=-76.6 kJ

(½) N2(g)+(3/2)O2(g)+(½)H2(g) → HNO3(l)

DH°=-174.1 kJ

Answer: 2 H2O(l) → 2 H2(g) + O2(g) DH°=+571.6 kJ 4 HNO3(l) → 2N2O5(g) + 2H2O(l) DH°=+153.2 kJ 2N2(g)+6O2(g)+2H2(g) → 4HNO3(l) DH°=-696.4 kJ

N2(g) + (5/2) O2(g) → N2O5(g)

DH° = +28.4 kJ

PROBLEM 10: Calculate DHf° for this reaction: 6 C(s) + 6 H2(g) + 3 O2(g) → C6H12O6(g)

using the following three equations: C(s) + O2(g) → H2(g) + (½) O2(g) →

CO2(g) DH° = -393.51 kJ H2O(l) DH° = -285.83 kJ

C6H12O6(s)+6 O2(g)→ 6CO2(g)+6 H2O(l) DH° = -2803.02 kJ

The answer is -1273.02 kJ/mol.

CONCLUSIONS • We practice different way to solve problems about Hees’s Law and Bond Energies associated with first Law of Thermodynamic. • Is very important be careful with the units that we use.