Chemistry 102 – Chapter 6 Summary and Review Answers 1. What are the definitions of system, surroundings, and universe? What is the first law of thermodynamics? Answer: Rather than give the textbook definitions for this here. Try this: write down what you think the definition is, then consult the textbook (within Chapter 6 rather than the glossary) and write this definition. Compare the two and determine if they are the same. If not, try and revise your definition to be correct but still in your own words. Give a diagram to assist this. 2. What is the relationship between heat and enthalpy? How is enthalpy different from internal energy? When is change in enthalpy equal to change in internal energy? Answer: Energy equals work and heat (or according to the first law of thermodynamics, the energy of a system is constant unless work is done on the system or is done by the system or heat is evolved or absorbed). So we have ∆E = w + q. Considering expansion or compression work (PV work), work is –P∆V. If the volume is constant, then ∆V = 0 and w = 0. Then ∆E = qV (constant volume). However, under constant pressure (a more common set of experimental conditions – open container under the pressure of the atmosphere), work is not zero (∆V ≠ 0). Now, if we define qP as a state function, enthalpy, ∆H = qP. In relation to internal energy, ∆E = ∆H – P∆V, or ∆H = ∆E + P∆V = ∆E + RT∆ngas; particularly for reactions involving ∆ngas ≠ 0, ∆E ≠ ∆H. Internal energy (as measured by change in energy entering or leaving the system) is heat and work; change in enthalpy is the heat measured at constant pressure. 3. When a hot sample of iron is placed in a constant pressure calorimeter containing cold water and assuming no loss of heat to the surroundings, what occurs with regards to the heat transfer between the iron and water? How does this translate to the initial temperature of the water/iron and the final temperature of the water and iron? Answer: The heat transfer between the iron and water has the thermal energy flowing from the iron to the water with the iron losing heat (energy) and the water gaining: –qiron = qwater The water was initially cold and because of the heat gained, is now warmer or: The iron was initially hot and because of the heat loss, is now colder or: Ti(water) < Tf(water) ∆Twater > 0 Ti(iron) > Tf(iron) ∆Tiron < 0 [Which is endothermic and which is exothermic? If this iron was allowed to cool by the surroundings, what is the temperature change of the surroundings? If the system is enothermic, what is the surroundings?] 4. Pictorially represent a constant pressure calorimeter. If 50 mL of 0.250 M sodium hydroxide and 50 mL of 0.250 M hydrochloric acid both initially at 25.00oC are combined (with no heat loss to the surroundings), what is the final temperature on the calorimeter? What is the calorimeter constant? Answer: A constant pressure calorimeter is represented to the right. The calorimeter constant can be approximated to be due to the water: q = C∆T; q = ms∆T (because water has a constant composition), so Cwater = mwaterswater = (100 g)(4.18 J∙g–1∙oC–1) = 418 J∙oC–1 The reaction (system) is the neutralization of sodium hydroxide with hydrochloric acid (expect this to be exothermic): HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq) Or the net ionic equation of: H+(aq) + OH–(aq) → H2O(l) Calculating the change in enthalpy for the equation: ΔH rxn = ∑ nproducts ΔH
( products ) − ∑ nreactants ΔH of ( reactants )
= ΔH of ( H 2 O ( l ) ) − ⎡⎣ ΔH of ( H + ( aq ) ) + ΔH of ( OH − ( aq ) ) ⎤⎦
o
f
= ( −285.8 kJ ) − ⎡⎣ 0 + ( −229.94 kJ ) ⎤⎦ = −55.9 kJ ( the reaction is exothermic as expected )
Therefore for the reaction of 50.0 mL of 0.250 M HCl and 50.0 mL of 0.250 M NaOH is the reaction of 0.0125 mol of H+ and 0.0125 mol of OH–. This makes 0.0125 mol of water. With –55.9 kJ of heat produced for every mol of water produced, for this reaction, –699 J of heat is produced (the product of the molar change in enthalpy and the amount for this reaction: –55.9 kJ∙mol–1 × 0.0125 mol). Therefore, qrxn = –699 J; –qrxn = qcal; qcal = +699 J qcal = Ccal ΔTcal
Tf =
ΔTcal =
qcal
= T f − Ti
Ccal
699 J
+ 25.0 o C = 26.7 o C
418 J ⋅ o C −1
Tf =
qcal
+ Ti
Ccal
5. Explain endothermic vs. exothermic in terms of heat gain/loss and the system; change in temperature of system/surroundings; enthalpy of the reactants and products and an energy diagram showing this. Answer: Heat loss / gain: Exothermic reactions (systems) lose heat or heat leaves the system. Endothermic reactions (systems) gain heat or heat enters the system. Change in temperature of the system/surroundings: Exothermic systems, the heat is lost from the system to the surroundings. Therefore, the temperature of the system is decreasing or change in temperature is negative and the temperature of the surroundings is increasing or change in temperature is positive. Endothermic systems, the heat is gained from the surroundings to the system. Therefore, the temperature of the system is increasing or change in temperature is positive and the temperature of the surroundings is decreasing or change in temperature is negative. Enthalpy of reactants and products: Exothermic systems, the change in enthalpy of the products must be less than the change in enthalpy of the reactants, such that: ∆Hproducts – ∆Hreactants < 0 Endothermic systems, the change in enthalpy of the products must be greater than the change in enthalpy of the reactants, such that: ∆Hproducts – ∆Hreactants > 0 Energy diagrams:
Endothermic Exothermic [What is the value of ∆Hrxn from these figures?] 6. What are the standard conditions for change in enthalpy? What is the thermochemical equation for the formation of methanol (CH3OH)? What is the standard enthalpy of combustion of methanol? How much heat is generated under standard conditions when 125 g of methanol are combusted? Answer: Formation of methanol: C(s, graphite) + 2H2(g) + ½O2(g) → CH3OH(l) ∆Hfo = –238.7 kJ∙mol–1 (p. A‐5) Combustion of methanol: CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(l) ΔH comb = ∑ nproducts ΔH
( products ) − ∑ nreactants ΔH of ( reactants )
= ΔH of ( CO 2 ) + 2ΔH of ( H 2 O ) − ΔH of ( CH 3 OH )
= −393.5 kJ + 2 ( −285.8 kJ ) − ( −238.7 kJ ) = −726.4 kJ
The standard conditions are partial pressure of all gas at 1 bar of pressure (1 bar = 0.987 atm, so very close to 1 atm), 298 K or 25oC and concentration of all solutions at 1 M. o
f
For 125 g of methanol, this is 3.90 mol so –2826 kJ is evolved.