Chemistry, Canadian Edition Chapter 03: Student Study Guide
Chapter 3: Energy and Its Conservation
Learning Objectives
Upon completion of this chapter you should be able to:
• recognize the types of energy of interest to chemists
• understand the first law of thermodynamics and the concepts of heat and work
• understand the origins of energy changes in chemical reactions
• apply the principles of calorimetry to determine energy changes in a chemical reaction
• understand and calculate enthalpy and internal energy
• be familiar with our sources of energy
Practical Aspects
The study of chemical energetics is one of the most practical everyday chemistry topics to study. A basic understanding of how energy flows is crucial to understanding most scientific phenomena. As you will learn, energy is composed of heat and work.
At its most basic level, chemists study chemical energetics to understand the relative strengths of chemical bonds. From a practical standpoint, chemists want to know whether or not a given chemical reaction will produce/release heat energy. An applications chemist would use the knowledge from this chapter to assess a given physical or chemical process and attempt to determine conditions which would yield the greatest amount of usable energy for that particular application. An engineer would be interested in constructing systems that would maximize either heat or work for a given process. Engineers would also want a conceptual understanding of why different materials absorb heat differently, so that they could choose the correct material for a given job. A biologist might be interested in studying the various caloric values of foods that provide our bodies with energy.
The concepts covered in this chapter will be used again in Chapter 11.
3.1 types of energy
Skill to Master:
· Identifying forms of energy.
Key Terms, Key Concepts, and Useful Relationships:
· These are summarized in a table on the next page.
Units and Conversions:
· Energy is measured in Joules (J).
· 1 J = 1 kg m2/sec2
Summary Table of Various Forms of Energy
Kinetic Energy, KE / Energy of Motion / KE = ½ mu2, where m = mass and u = speed
Potential Energy, PE / Energy of Position,
Stored Energy / Both definitions of PE are used frequently in chemistry.
Chemical Energy / Energy stored inside a molecule resulting from attractive forces between the atoms within the molecule
(a type of potential energy) / When a substance undergoes a chemical transformation, chemical energy is either released or absorbed by the substance.
Thermal Energy / Constitutes a variety of energy types, depending on the substance analyzed: kinetic, rotational, vibrational, translational. Heating an object increases these energies. / Example: Kinetic energy of a monoatomic gas, as seen in Chapter 2:
Electrical Energy / Energy arising from forces between two charged particles
(a type of potential energy) /
where k = 2.31x10-16 J pm, q1 and q2 = charges on the respective objects, r = distance between the objects.
Radiant Energy / Light Energy / UV light, visible light, Infrared light are some examples.
(covered in detail in 7)
Exercise 1: List the energy transformations that occur during each process: a) moving a notebook from the floor to a high shelf; b) providing energy to heat your home with an electrical heater.
Solution:
a) Kinetic energy (moving the notebook) is converted into potential energy (the notebook now has potential to fall to the floor). Technically, chemical energy was transformed into the kinetic energy because chemical reactions had to take place inside your body to provide you with energy to move.
b) Electrical energy from a power plant (or several power plants) is converted into thermal energy in your home. The electrical energy could have come from a variety of sources: 1) a waterfall or a windmill – kinetic energy from the moving water or wind gets converted to electrical energy; 2) a nuclear power plant would convert nuclear chemical energy into thermal energy by heating up water to make steam, which would be converted to kinetic energy by making the steam move a turbine, which would be converted to electrical energy; 3) a coal plant or other fuel plant would convert chemical energy into thermal energy then kinetic energy, then electrical energy.
Try It #1: List the energy transformations that occur when you burn a log in a campfire.
Exercise 2: Determine the kinetic energy of an electron traveling at 95 kilometres per hour.
Strategy: Use the equation: “KE = ½ mu2” to determine the kinetic energy. The value for speed will have to be in units of “m/sec;” we can do that first or incorporate it into the problem. Both methods will be shown to illustrate that the same result is obtained.
Method 1: Convert rate units to m/sec first, then apply that to the KE equation in a second step.
KE = ½
Method 2: Do calculation set-up and conversion in one step.
KE = ½
Notice in the second method that the conversion factors had to be squared in order for the units to cancel out. When punching numbers into a calculator, be sure to account for the squared terms.
Solution: The kinetic energy of one electron traveling at 95 km/hour is 3.1x10-28 J. (2 sig figs from data)
Take-home messages from Exercise 2:
· There’s often more than one correct way to approach a chemistry problem.
· When doing a multi-step mathematical process – as in Method 1 – wait until the end of the last step to round off to sig figs.
3.2 thermodynamics
Skills to Master:
· Calculating the heat that flows between the system and the surroundings.
· Calculating the work done on or by the system.
· Applying the first law of thermodynamics.
Thermodynamics is the study of energy flow. Energy flows from one object to another, but is never created or destroyed. In order to discuss and quantitate energy flow, several terms must be defined:
Key Terms:
· Law of Conservation of Energy – Energy cannot be created nor destroyed. It can only be transferred from one body to another or converted from one form to another.
· System – the material(s) being studied.
· Surroundings – everything else in the universe except for the system.
· Boundary – that which separates the system from the surroundings.
Key Concept:
· Depending upon what you want to analyze, you might define a particular system differently than another person. The “system” is simply “what is being studied.”
Exercise 3: Let’s say we’d like to study the thermodynamics of a cup of coffee. Define the system, surroundings, and boundary in this situation.
Solution: We would define the system as the coffee itself (i.e., the contents of the cup). The boundary would be the cup, and the surroundings would be every other thing in the entire universe.
Heat and Energy
Key Terms:
· Temperature (T) - measure of kinetic energy of molecular motion.
· Heat (q) – transfer of thermal energy between a system and its surroundings.
· Molar heat capacity (C) – amount of heat energy required to raise one mole of substance by one degree Celsius.
Key Concepts:
· Heat always flows from hot to cold.
· DT depends upon four factors:
1) amount of heat transferred – the more heat transferred, the greater the DT;
2) direction of heat transferred – DT = Tf – Ti, so initial vs. final T will predict sign of q;
3) amount of material – the more substance present, the more heat it can absorb or release;
4) identity of material – some materials absorb heat better than others.
· DT in units of Kelvin is the same as DT in units of degrees Celsius.
For example: raising the temperature from 10°C to 25°C gives a DT of +15°C.
In Kelvin it would be: (25 + 273) – (10 + 273) = +15 K.
· When energy flows into a system from the surroundings, the system gains energy.
· When energy flows from a system into the surroundings, the system loses energy.
· Energy changes are measured from the point of view of the system. If no subscript is noted then the value is referring to the system.
Helpful Hints
· Metals have low heat capacities; they are good conductors of heat.
· Water has an unusually high heat capacity.
Useful Relationships:
· q = nCDT
· qsurroundings = -qsystem; heat flows between the system and the surroundings, so their values will be equal in magnitude, but opposite in sign.
Exercise 4: If you have 10 moles each of copper and water – one is very hot and the other is very cold – and you simultaneously toss them into the same insulated container,
a) which will gain or lose more heat?
b) what will be true of the magnitudes of the DTs?
c) If the copper was originally the hotter object, what will be the sign of DT for each substance?
d) If the copper is considered the system and the water is considered the surrounding, what is the relationship between q for each of these?
Strategy: Assess what is occurring. The two substances are initially at different temperatures, so when they come in contact, heat will flow from one to the other until their temperatures are equal.
Solution:
a) The heat lost by one will equal the heat gained by the other. Neither will gain nor lose more than the other.
b) DTCu will not equal because the substances have different molar heat capacities. Copper, with a lower heat capacity, will undergo a greater temperature change than water.
c) Cu: temperature is decreasing, so Tf is smaller than Ti. The sign for DTCu will be negative. The sign for DTwater will be positive.
d) q for copper will be equal in magnitude but opposite in sign to q for water. We could write this as: qcopper = - qwater or -qcopper = qwater. Either equation is valid; notice that both equations simply illustrate these values are opposite in sign from each other.
Exercise 5: A 2.0 kg cast iron skillet at room temperature (19°C) is placed in an oven at 175°C. How much heat energy does the skillet absorb when it heats up to the oven’s temperature? Assume that the skillet is pure iron for the calculation. Ciron is 25.10 J/mol K.
Strategy: We’ll use a shortcut of the 7-step method to analyze this problem. We’re asked to find qiron. We’ve been given the mass of iron and its temperature change. We can use “q=nCDT.” Here are some considerations:
· DT is always final minus initial conditions: 175° – 19° = 156°C; DT = 156 K.
· Mass can be converted to moles using molar mass.
Solution: The heat energy required is 1.4x105 J. This answer seems reasonable because it is a positive number, which indicates heat was absorbed by the iron.
Try It #2: If a 0.46 kg hot copper pan lost 9.8x103 J of heat as it cooled down to 54°C, what was its original temperature?
Work
Key Term:
· Work (w) – energy used to move an object against an opposing force.
Key Concepts:
· When a system does work on the surroundings, the system loses energy, so the sign of w is negative.
· When the surroundings do work on a system, the system gains energy, so the sign of w is positive.
· Like heat, work is transferred between the system and the surroundings, so work of the system will be equal in magnitude but opposite in sign from work of the surroundings.
Useful Relationships:
· wsurroundings = -wsystem
First Law of Thermodynamics: Law of Conservation of Energy
Key Term:
· Law of Conservation of Energy (also known as the First Law of Thermodynamics) – Energy cannot be created nor destroyed, just transferred. (In other words, the energy of the universe is constant.)
Key Concepts:
· The only two types of energy that can be exchanged between a system and its surroundings are heat (q) and work (w): DE = q + w.
· The energy gained by the system is lost by the surroundings and vice versa.
Useful Relationships:
· The relationship: DE = q + w can be incorporated with what we already know about heat, work and energy to give us many useful relationships. We use the subscripts “sys” and “surr” for “system” and “surroundings”, respectively, to keep track. Convince yourself that these relationships are true:
o -DEsurr = qsys + wsys
o DEsys + DEsurr = DEtotal = 0
o qsys = DEsys - wsys
Exercise 6: If a system does 15 J of work and receives 18 J of heat, what is the value of DE for the system?
Strategy: The key to answering this question is to remember what the signs for q and w indicate: if the system receives heat, the value for q is “+,” and if it does work the value for w is “-.”
Solution: DEsys = qsys + wsys = (18 J) + (-15 J) = 3 J (1 sig fig from subtraction rules)
State Functions and Path Functions
Key Terms:
· State – the conditions that describe the system (For example: pressure, volume, temperature, amount of substance).
· Change of state – a change in conditions of the system.
· State function – a function or property whose value depends only on the present state (condition) of the system, not on the path used to arrive at that condition.
· Path function – a function that depends upon how the change occurs.
Important features of state functions:
· A state function can be analyzed from its initial and final conditions. The process in going from the initial to final conditions does not need to be known.
· For any state function, the overall change is zero if the system returns to its original condition.
· Reversibility: if a state function has a numerical value of “+5” in one direction, it will have a value of “-5” in the reverse direction.
· Values for many state functions are tabulated in reference tables and books.