Using Hess’s Law and Bond Energy to calculate enthalpy change. IB-Chemistry

In-Class Assignment

1. Given the enthalpy changes for the reactions below

2 H2O2 (aq) 2 H2O (l) + O2 (g) ΔH = -200 kJ/mol

2 H2 (g) + O2 (g) 2 H2O (l) ΔH = -600kJ/mol

what will be the enthalpy change for : H2 (g) + O2 (g) H2O2 (aq) ?

A. -200 kJ/mol B. -400 kJ/mol C. -600 kJ/mol D. -800 kJ/mol

2. Iron and chlorine react directly to form iron(III) chloride, not iron(II) chloride. Therefore it is not possible to directly measure the enthalpy change for the reaction

Fe(s) + Cl2 (g) FeCl2 (s)

The enthalpy changes for the formation of iron(III) chloride from the reaction of chlorine with iron and with iron(II) chloride are given below. Use these to calculate the enthalpy change for the reaction of iron with chlorine to form iron (II) chloride.

2 Fe (s) + 3 Cl2 (g) 2 FeCl3 (s) ΔH = -800 kJ/mol

2 FeCl2 (s) + Cl2 (g) 2 FeCl3 (s) ΔH = -120 kJ/mol

3. The Romans used calcium oxide (CaO) as mortar in stone structures. The CaO was mixed with water to give Ca(OH)2, which slowly reacted with CO2 in the air to give limestone.

Ca(OH)2 (s) + CO2 (g) CaCO3 (s) + H2O (l)

Use the enthalpies of the following reactions to calculate the enthalpy change for the reaction above.

Ca (s) + O2 (g) + H2 (g) Ca(OH)2 (s) ΔH = -986.1 kJ/mol

C (s) + O2 (g) CO2 (g) ΔH = -394.1 kJ/mol H2(g) + 1/2 O2(g) H2O(l) ΔH = -285.8 kJ/mol Ca (s) + C (s) + 3/2 O2 (g) CaCO3(s) ΔH = -1206.9 kJ/mol

4. Which of the following equations is equivalent to the bond enthalpy of the carbon-oxygen bond in carbon monoxide?

A. CO (g) C(g) + O (g) B. CO (g) C(s) + O (g) C. CO (g) C(s) + ½ O2 (g) D. CO (g) C(g) + ½ O2 (g)

5. The bond enthalpy of the N-O bond in nitrogen dioxide is 305 kJ/mol. If that of the bonds in the oxygen molecule and the nitrogen molecule are 496 kJ/mol and 944 kJ/mol respectively, what will be the enthalpy change for the following reaction?

N2 (g) + 2 O2 (g) 2 NO2 (g)

A. 716 kJ/mol B. 1135 kJ/mol C. 1326 kJ/mol D. 1631 kJ/mol

6. The enthalpy change for the following reaction is +688 kJ/mol. N2 (g) + 3 Cl2 (g) 2 NCl3 (g)

Calculate the bond enthalpy of the N-Cl bond, given that the bond enthalpies of the nitrogen molecule and the chlorine molecule are 944 kJ/mol and 242 kJ/mol respectively.

7. Use bond energy data to calculate the enthalpy change when cyclopropane reacts with hydrogen to form propane. The actual value is –159 kJ/mol. Give the reasons why you think this differs from the value you have calculated.

[Bond energies in kJ/mol: C-C = 348; C-H = 412; H-H = 436]

The reaction is shown in the following chemical equation:

CH2

CH2CH2+ H2

CH3 CH2

CH3