Energy and Rates of Reaction - Lesson 3 - Homework Answers

Heats and Rates of Reaction – Lesson 3 - Homework

Use the Bond Energy Table from Lesson 3 to answer all of the following problems. 1. Hydrogenation of double and triple bonds is an important industrial process. Calculate (in kJ) the standard enthalpy change ΔH for the hydrogenation of ethyne (acetylene) to ethane: H−C≡C−H(g) + 2H2(g)  H3C−CH3(g) ΔHrxn =

[1 mol C≡C * (839 kJ/mol C≡C) + 2 mol H-C * (413 kJ/mol H-C) + 2 mol H-H * (432 kJ/mol H-H)] – [6 mol H-C * (413 kJ/mol H-C) + 1 mol C-C * (347 kJ/mol C-C)]

ΔHrxn = - 296 kJ

2. Using bond enthalpies, calculate the reaction enthalpy (ΔH) for: CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) ΔHrxn =

[4 mol C-H * (413 kJ/mol C-H) + 1 mol Cl-Cl * (239 kJ/mol Cl-Cl)] – [3 mol C-H * (413 kJ/mol C-H) + 1 mol C-Cl * (339 kJ/mol C-Cl) + 1 mol H-Cl * (427 kJ/mol H-Cl)]

ΔHrxn =-114 kJ

3. What is the enthalpy of reaction for the following equation: 2 CH3OH(l) + 3 O2(g)  2 CO2(g) + 4 H2O(g) ΔHrxn =

[6 mol C-H * (413 kJ/mol C-H) + 2 mol C-O * (358 kJ/mol C-O) + 2 mol H-O * (467 kJ/mol H-O) + 3 mol O=O * (495 kJ/mol O=O)] - [4 mol C=O * (799 kJ/mol C=O) + 8 mol H-O * (467 kJ/mol H-O)]

ΔHrxn = - 1319 kJ

4. The reaction of H2 with F2 produces HF with ΔH = -269 kJ/mol of HF. If the H-H and H-F bond energies are 432 and 565 kJ/mol, respectively, calculate the F-F bond energy? How does this compare to the value on the table? H2(g) + F2(g)  2HF(g) ΔHrxn

= -269 kJ/mol HF * 2 mol HF = - 538 kJ

- 538 kJ =

[1 mol H-H * (432 kJ/mol H-H) + 1 mol F-F * (Bond Energy F-F)] - [2 mol H-F * (565 kJ/mol H-F)]

Therefore Bond Energy of F-F = 160 kJ/mol of F-F This is similar to the table value of 154 kJ/mol.

5. Determine the enthalpy of the following reaction: CH3CH=CH2 + 9/5 O=O  3 O=C=O + 3 H-O-H ΔHrxn =

[1 mol C-C * (347 kJ/mol C-C) + 1 mol C=C * (614 kJ/mol C=C) + 6 mol C-H * (413 kJ/mol C-H) + 9/5 mol O=O * (495 kJ/mol)] - [6 mol C=O * (799 kJ/mol C=O) + 6 mol H-O * (467 kJ/mol H-O)]

ΔHrxn = - 3266 kJ

6. Calculate ΔH for this reaction: H2C=CH2(g) + H2O(l)  CH3-CH2-OH(l) ΔHrxn =

[1 mol C=C * (614 kJ/mol C=C) + 4 mol C-H * (413 kJ/mol C-H) + 2 mol H-O * (467 kJ/mol H-O)] - [5 mol C-H * (413 kJ/mol C-H) + 1 mol C-C * (347 kJ/mol C-C) + 1 mol C-O * (358 kJ/mol C-O) + 1 mol H-O * (467 kJ/mol H-O)]

ΔHrxn =

- 37 kJ