Part II. Multiple choice, continued (3 points each). Name:__________________ Please circle your answer. There is only one correct answer to each question. 22. ∆E for a system that releases 12.4 J of heat and does 4.2 J of work on the surroundings is ____________ J. A) 16.6 B) 12.4 C) –8.2 D) –16.6 23. ∆H for an endothermic process is _________ while ∆H for an exothermic process is __________. A) zero, positive B) positive, negative C) negative, zero D) negative, positive 24. The value of ∆H˚ for the reaction below is –283 kJ. Calculate the heat (kJ) released to the surroundings when 12.0 g of CO reacts completely. CO (g) + O2 (g) à CO2 (g) A) 121 B) 660 C) 283 D) 212 25. Given the following reactions N2 (g) + 2O2 (g) à 2NO2 (g) ∆H = 66.4 kJ 2NO (g) + O2 (g) à 2NO2 (g) ∆H = -114.2 kJ the enthalpy of the reaction of nitrogen to produce nitric oxide N2 (g) + O2 (g) à 2NO (g) is __________ kJ. A) 180.6 B) –47.8 C) 47.8 D) –180.6 26. Given the data in the table below, ∆H˚ for the reaction 2CO (g) + O2 (g) à 2CO2 (g) is __________ kJ. Substance ∆Hf˚ (kJ/mol) CO (g) -110.5 CO2 (g) -393.5 CaCO3 (s) -1207.1 A) –677.0 B) 283.3 C) –283.0 D) –566.0

Here's an in-depth explanation of the concepts behind those multiple-choice questions:

22. ∆E for a system that releases 12.4 J of heat and does 4.2 J of work on the surroundings is J.
Answer: D) –16.6
Explanation:

• The change in internal energy (\Delta E) of a system is given by the first law of thermodynamics: \Delta E = q + w, where q is heat and w is work.
• When heat is released by the system, q is negative. So, q = -12.4 J.
• When the system does work on the surroundings, w is negative. So, w = -4.2 J.
• Therefore, \Delta E = -12.4 + (-4.2) = -16.6 J

23. ∆H for an endothermic process is _ while ∆H for an exothermic process is __.
Answer: B) positive, negative
Explanation:

• Endothermic process: A process that absorbs heat from the surroundings. Since the system gains heat, the enthalpy change (\Delta H) is positive.
• Exothermic process: A process that releases heat to the surroundings. Since the system loses heat, the enthalpy change (\Delta H) is negative.

24. The value of ∆H˚ for the reaction below is –283 kJ. Calculate the heat (kJ) released to the surroundings when 12.0 g of CO reacts completely.
CO (g) + O2 (g) → CO2 (g)
Answer: A) 121
Explanation:

• First, find the number of moles of CO: The molar mass of CO is approximately 28.0 g/mol. So, n = \frac{12.0 \text{ g}}{28.0 \text{ g/mol}} = 0.429 mol.
• The given reaction shows that when 1 mole of CO reacts, 283 kJ of heat is released. Therefore, for 0.429 mol of CO, the heat released is:
  q = 0.429 \text{ mol} \times 283 \text{ kJ/mol} = 121.407 \text{ kJ}
• So, approximately 121 kJ of heat is released.

25. Given the following reactions
N2 (g) + 2O2 (g) → 2NO2 (g) \Delta H = 66.4 kJ
2NO (g) + O2 (g) → 2NO2 (g) \Delta H = -114.2 kJ
the enthalpy of the reaction of nitrogen to produce nitric oxide
N2 (g) + O2 (g) → 2NO (g)
is __ kJ.
Answer: A) 180.6
Explanation:

• We need to find the enthalpy change for the reaction: N2 (g) + O2 (g) → 2NO (g).
• We can use Hess's Law to manipulate the given reactions:
  1. Reverse the second reaction: 2NO2 (g) → 2NO (g) + O2 (g) \Delta H = +114.2 kJ
  2. Keep the first reaction as is: N2 (g) + 2O2 (g) → 2NO2 (g) \Delta H = 66.4 kJ
• Add the two reactions:
  \text{N2 (g) + 2O2 (g) + 2NO2 (g) } \rightarrow \text{ 2NO2 (g) + 2NO (g) + O2 (g)}
  \text{N2 (g) + O2 (g) } \rightarrow \text{ 2NO (g) }
• Add the enthalpy changes: \Delta H = 66.4 + 114.2 = 180.6 kJ

26. Given the data in the table below, ∆H˚ for the reaction
2CO (g) + O2 (g) → 2CO2 (g)
is __ kJ.

Answer: D) –566.0
Explanation:

• Use the formula: \Delta H{rxn}^\circ = \sum \Delta H{f,products}^\circ - \sum \Delta H_{f,reactants}^\circ
• For the reaction 2CO (g) + O2 (g) → 2CO2 (g):
  \Delta H{rxn}^\circ = [2 \times \Delta Hf^\circ (CO2)] - [2 \times \Delta Hf^\circ (CO) + \Delta Hf^\circ (O2)]
• Since the standard enthalpy of formation of an element in its standard state is zero, \Delta Hf^\circ (O2) = 0
• \Delta H_{rxn}^\circ = [2 \times (-393.5)] - [2 \times (-110.5) + 0]
• \Delta H_{rxn}^\circ = -787 + 221 = -566 kJ