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Chemistry Textbook

Energy Basics and Enthalpy

By the end of this section, you will be able to:

  • Define energy, distinguish types of energy, and describe the nature of energy changes that accompany chemical and physical changes

  • State the first law of thermodynamics

  • Define enthalpy and explain its classification as a state function

This chapter will introduce the basic ideas of an important area of science concerned with the amount of heat absorbed or released during chemical and physical changes—an area called thermochemistry. The concepts introduced in this chapter are widely used in almost all scientific and technical fields. Food scientists use them to determine the energy content of foods. Biologists study the energetics of living organisms, such as the metabolic combustion of sugar into carbon dioxide and water. The oil, gas, and transportation industries, renewable energy providers, and many others endeavor to find better methods to produce energy for our commercial and personal needs. Engineers strive to improve energy efficiency, find better ways to heat and cool our homes, refrigerate our food and drinks, and meet the energy and cooling needs of computers and electronics, among other applications. Understanding thermochemical principles is essential for chemists, physicists, biologists, geologists, every type of engineer, and just about anyone who studies or does any kind of science. 

Energy

Energy can be defined as the capacity to supply heat or do work. One type of work (w) is the process of causing matter to move against an opposing force. For example, we do work when we inflate a bicycle tire—we move matter (the air in the pump) against the opposing force of the air already in the tire.

Like matter, energy comes in different types. It is an elusive driving force for all physical and chemical changes, and exists in many forms such as heat, mechanical work (including kinetic and potential energy), light (radiation), sound, electric, and chemical energy. Kinetic energy, the energy that an object possesses because of its motion, and potential energy, the energy an object has because of its relative position, composition, or condition. Water at the top of a waterfall or dam has potential energy because of its position; when it flows downward through generators, it has kinetic energy that can be used to do work and produce electricity in a hydroelectric plant (Figure 5.2). Kinetic energy associated with the random motion of atoms and molecules is the thermal energy possessed by the enitity.

E Kinetic= 12 mv2

"m" is the mass of the object expressed in kg, "v" is the velocity (in m/s) with which the object is moving.

E Potential=mgh

"h" is the height and "g" is the acceleration due to gravity (9.8 m/s² for Earth's surface)

Two pictures are shown and labeled a and b. Picture a shows a large waterfall with water falling from a high elevation at the top of the falls to a lower elevation. The second picture is a view looking down into the Hoover Dam. Water is shown behind the high wall of the dam on one side and at the base of the dam on the other.
Figure 5.2 (a) Water at a higher elevation, for example, at the top of Victoria Falls, has a higher potential energy than water at a lower elevation. As the water falls, some of its potential energy is converted into kinetic energy. (b) If the water flows through generators at the bottom of a dam, such as the Hoover Dam shown here, its kinetic energy is converted into electrical energy. (credit a: modification of work by Steve Jurvetson; credit b: modification of work by “curimedia”/Wikimedia commons)

In the electrical case, a charge will exert a force on any other charge and potential energy arises from any collection of charges. A battery has electric potential energy because the chemicals within it can produce electricity that can do work. For example, if a positive charge Q is fixed at some point in space, any other positive charge which is brought close to it will experience a repulsive force and will therefore have potential energy. The potential energy of a test charge q in the vicinity of this source charge will be: 

E electric potential=k Qqr

k, is Coulomb's constant and r is the distance between the point charges.

The Principle of Conservation of Energy

Energy is transmitted in the form of heat from one place to another or in the form of mechanical work (potential and kinetic energy). Both types of transmission need a medium. Heat is usually released or absorbed, but sometimes the conversion involves light, electrical energy, or some other form of energy. At the atomic and molecular level, transmission of heat is also a result of transferring of kinetic energy among atoms, molecules or ions in the medium. Transmission of energy without any medium is a phenomena known as electromagnetic radiation, in which bundles of energy are emitted as photons of light according to Max Planck. 

Energy can be used to perform mechanical work and is required to cause any change, physical or chemical. During the transformation from one form to another, amounts of energy remain the same. Energy cannot be destroyed or created. This is the principle of conservation of energy.

To understand the principle of conservation of energy in the energy transfer processes, we have to isolate a system from its surroundings. System is a quantity of matter or a region in space chosen for study. For example, a closed container with gas or liquid in it is a system; so is a machine. Surroundings is the region outside the system. 

Internal Energy and Enthalpy

Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy (U), sometimes symbolized as E.

As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (q) from the surroundings or when the surroundings do work (w) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire’s temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.

The relationship between internal energy, heat, and work can be represented by the equation:

Δ U = q + w Δ U = q + w
 

as shown in Figure 5.3. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive q is heat flow in; negative q is heat flow out) or work done on or by the system. The work, w, is positive if it is done on the system and negative if it is done by the system.

A rectangular diagram is shown. A green oval lies in the center of a tan field inside of a gray box. The tan field is labeled “Surroundings” and the equation “Δ U = q + w” is written at the bottom of the diagram. Two arrows face into the green oval and are labeled “q subscript in” and “w subscript on” while two more arrows face away from the oval and are labeled “q subscript out” and “w subscript by.” The center of the oval contains the terms “Δ U > 0”, “System,” and “Δ U < 0.”
Figure 5.3  The internal energy, U, of a system can be changed by heat flow and work. If heat flows into the system, qin, or work is done on the system, won, its internal energy increases, ΔU> 0. If heat flows out of the system, qout, or work is done by the system, wby, its internal energy decreases, ΔU < 0.

A type of work called expansion work (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.

Chemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system’s internal energy (U) and the mathematical product of its pressure (P) and volume (V):

H=U+PVH=U+PV

Enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change (ΔH) is:

Δ H = Δ U + P Δ V Δ H = Δ U + P Δ V

The mathematical product PΔV represents work (w), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of ΔV and w will always be opposite:

PΔV=−wPΔV=−w

Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:

Δ H = Δ U + P Δ V = q p + w − w = q p ΔH=ΔU+PΔV=qp+w−w=qp

where qp is the heat of reaction under conditions of constant pressure.

And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (qp) and enthalpy change (ΔH) for the process are equal.

 

EXAMPLE 5.1
Calculating Kinetic Energy

If the mass of a baseball is 149 g, what is the kinetic energy of a fastball clocked at 100 mi/h?

Solution:

E (Kinetic Energy) = (½) mv²

In this case, we know both mass and velocity, however, they must be converted to respective SI units. 

 v = ( 100 m i 1 h )   ( 1 h 60 m i n )   ( 1 m i n 60 s ) ( 1.61 k m 1 m i )   ( 1000 m 1 k m ) = 44.7 m / s

So, kinetic energy is:

K E = 149 g ( 1 k g 1000 g ) ( 44.7 m s ) 2 = 1.49 × 10 2   k g ⋅ m 2 s 2 = 1.49 × 10 2 J
Check your Learning:

In a bowling alley, the distance from the foul line to the head pin is 59 ft, 10 13/16 in. (18.26 m). If a 16 lb (7.3 kg) bowling ball takes 2.0 s to reach the head pin, what is its kinetic energy at impact? (Assume its speed is constant.)

Answer:
3.10× 10² J
EXAMPLE 5.2
Calculating Potential Energy

A batter hits a pop fly, and the baseball (with a mass of 149 g) reaches an altitude of 250 ft. If we assume that the ball was 3 ft above home plate when hit by the batter, what is the increase in its potential energy?

Solution:

The increase in potential energy is the same as the amount of work required to raise the ball to its new altitude, which is (250 − 3) = 247 feet above its initial position. Thus, potential energy is:

P E = 149 g ( 1 k g 1000 g ) ( 9.81 m s 2 ) ( 247 f t ) ( 0.3048 m 1 f t ) = 1.10 × 10 2   k g ⋅ m 2 s 2 = 1.10 × 10 2 J
Check your Learning:

What is the potential energy of a 16 lb bowling ball held 3.0 ft above your foot?

Answer:

65 J

EXAMPLE 5.3
Example 5.3: Finding internal energy

A cylinder and piston assembly is warmed by an external flame. The contents of the cylinder expand, doing work on the surroundings by pushing the piston outward against the external pressure. If the systems absorbs 559 J of heat and does 488 J of work during expansion, what is the value of ΔU?

Solution:

System absorbing heat, which means q = 559 J.

Work done on the surroundings = –488 J

So,  ΔU = q + w 

∆U=559J+-488J=71 J

 

Check your Learning:

A system releases 622 kJ of heat and does 105 kJ of work on the surroundings. What is the change in internal energy of the system?

Answer:

Answer: -7.27 × 10² kJ

EXAMPLE 5.4
Finding internal energy from pressure and volume

A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules?

Solution:

We know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention.

Because the gas is being compressed, we know that work is being done on the system, so w must be positive. 

w = − P ext Δ V = − 8.00  atm ( 0.0500 L − 0.400 L ) ( 101.3 J L ⋅ a t m ) = 284  J

Thus

ΔU = q + w = −140 J + 284 J = 144 J

In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs.

Check your Learning:

A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules?

Answer:

Answer: −216 J

EXAMPLE 5.5
Finding enthalpy using ideal gas equation

The combustion of graphite to produce carbon dioxide is described by the equation: C (graphite, s) + O2(g) → CO2(g). At 298 K and 1.0 atm, ΔH = −393.5 kJ/mol of graphite for this reaction, and the molar volume of graphite is 0.0053 L. What is ΔU for the reaction?

Solution:

In this reaction, 1 mol of gas (CO2) is produced, and 1 mol of gas (O2) is consumed. Thus Δn = 1 − 1 = 0. Substituting this calculated value and the given values into

Δ U = Δ H − R T Δ n = ( − 393.5 k J / m o l ) − [ 8.314 J / m o l ⋅ K ] ( 298 K ) ( 0 ) = ( − 393.5 k J / m o l ) − ( 0 J / m o l ) = − 393.5 k J

 

Check your Learning:

Calculate ΔU for the conversion of oxygen gas to ozone at 298 K: 3O2(g) → 2O3(g). The value of ΔH for the reaction is 285.4 kJ.

Answer:

288 kJ

Key Equations:

Δ U = q + w Δ U = q + w Δ H = Δ U + P Δ V Δ H = Δ U + P Δ V