Energy Relationships in Chemical Reactions

Energy Relationships in Chemical Reactions

Chapter 6

Energy Relationships in Chemical Reactions

The Nature of Energy and Types of Energy:

Energy: is the capacity to do work.

Work: directed energy change resulting from a process.

The total amount of energy = kinetic energy + potential energy

Kinetic Energy: the energy produced by a moving object. K.E. = ½ mv2 where m is the mass of the object and v is its velocity.

Potential Energy (P.E.): is the energy available by virtue of an object’s position.

Or is the energy an object has because it is either attracted to or repelled by some other object.

(P.E. = 0 if no attraction or repulsion)

  • When objects that attract each other are pulled apart, their P.E increases, and when they move toward each other, their P.E decreases.
  • When objects that repel are pushed toward each other, their P.E increases, and when they are moved apart, their P.E decreases.

Radiant Energy (Solar Energy): the energy coming from the sun as radiations.

Thermal energy: is the energy associated with the random motion of atoms and molecules.

Chemical energy: is the energy stored within the structural units (the bonds) of chemical substances

Nuclear energy: is the energy stored within the collection of neutrons and protons in the atom.

Electrical energy: is the energy associated with the flow of electrons.

All forms of energy can be converted from one form to another.

Law of Conservation of Energy: the total quantity of energy in the universe is assumed constant.

Energy Changes in Chemical Reactions:

Almost all chemical reactions absorb or release energy, generally in the form of heat.

Heat: is the transfer of thermal energy between two bodies that are at different temperatures.

Thermochemistry: the study of heat changes in chemical reactions.

System: specific part of the universe that is of interest to us.

Surroundings: every thing else other than the system. Or the rest of the universe outside the system.

Open System: can exchange mass and energy (heat) with its surroundings.

Closed System: which allows the transfer of energy

(heat) but not mass.

Isolated system: which does not allow the transfer of either mass or energy.

Exothermic Process: heat flow out to the surroundings.

Or any process that gives off heat (that is, transfers thermal energy to the surroundings).

The temperature of the reaction mixture rises and the P.E. of the chemicals involved in the reaction decreases.

2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) + energy

Endothermic Process: heat flow into the reaction mixture.

Or any process in which heat has to be supplied to the system from the surroundings.

The temperature of the reaction mixture drops and the P.E. of the chemicals involved in the reaction increases.

Energy + 2HgO(s)  2Hg(l) + O2(g)

Introduction to Thermodynamics:

Thermodynamics: The scientific study of the interconversion of heat and other kinds of energy.

State of a System: Defined by the values of all relevant macroscopic properties (which can be determined directly).e.g.: Composition, Energy, Temp., Pressure and Volume.

State Functions: are properties that are determined by the state of the system, regardless of how that condition was achieved.

Or variable is dependent on the current state of the system, and not on the path used to obtain the current conditions.

State functions as energy, enthalpy, volume, pressure, and temperature.

Non-state functions as heat and work.

Consider a gas at 2 atm, 300 K, and 1 L (the initial state), and the final state then corresponds to 1 atm, 300 K, 2 L. The change in volume (V) is

V = Vf - Vi = 2 L - 1 L = 1 L

No matter how we arrive the final state (for example, the pressure of the gas can be increased first and then decreased to 1 atm).

The 1st law of Thermodynamics: energy can be converted from one form to another, but cannot be created or destroyed. “The Total Energy of the Universe is Constant”.

We test the validity of the 1st law by measuring only the change in the internal energy (E) of a system between the initial state and the final state in a process.

E = Ef - Ei

If Ef = Ei then E = 0

Internal energy, E is a state function that corresponds to the total energy of the system (kinetic energy and potential energy).

Kinetic energy consists of various types of molecular motion and movement of electrons within molecules.

Potential energy is determined by the attractive interactions between electrons and nuclei and by repulsive interactions between electrons and between nuclei in individual molecules, as well as by interaction between molecules.

Consider the following reaction:

S(s) + O2(g)  SO2(g)

E = E(product) – E(reactants)

E = energy content of 1 mol of SO2(g) – [energy content of 1 mol of S(s) + 1 mol of O2(g)]

We find that this reaction gives off heat. Therefore, the energy of the product is less than that of the reactants, and E is negative.

The transfer of energy from the reactants to the surroundings does not change the total energy of the universe. That is, the sum of the energy changes must be zero:

Euniv = Esys + Esurr = 0or Esys = - Esurr

Energy gained in one place must have been lost somewhere else. Furthermore, because energy can be changed from one form to another, the energy lost by one system can be gained by another system in a different form.

Internal energy of a system changes in two ways: E = q + w

Eis the change in internal energy of a system.

q: heat exchange between the system and the surroundings.

w: work done on (or by) the system.

Sign Conventions for work and Heat
Process / Sign
Work done by the system on the surroundings / -
Work done on the system by the surroundings / +
Heat absorbed by the system from the surroundings (endothermic process) / +
Heat absorbed by the surroundings from the system (exothermic process) / -

If q and w are positive, energy enters the system, if q and w are negative, energy leaves the system.

Example: In a specific change, a system absorbs 50 J and does 30 J of work. Find E for the system and surroundings.

E = q + w

E = (+50) + (-30) = 20 J

Therefore, the system gained 20 J of energy. Therefore, Esurr = -20 J because Esys + Esurr = 0

q and w are notState Functions.

Work and Heat:

w = Fd

The work done by the gas on the surroundings is: w = -PVwhere V is given by Vf - Vi.

  • For gas expansion, V > 0, so -PV is a negative quantity.
  • For gas compression (work done on the system), V < 0, so -PV is a positive quantity.

The work done depends on the magnitude of the external, opposing pressure P. If P is zero (that is, if the gas is expanding against a vacuum), the work done must also be zero. If P is some positive, nonzero value, then the work done is given by -PV.

1 L . atm = 101.3 J

Example 6.1

Example 6.1 shows that work is not a state function. Although the initial and final states are the same in (a) and (b), the amount of work done is different because the external, opposing pressures are different.

We cannot write w = wf- wi for a change.

Work done depends not only on the initial state and the final state, but also on how the process is carried out.

Heat (q) is not a state function.

Suppose that the change can be brought about in a system in two different ways.

1st case: E = q1 + w1 = q1 since w1 = 0

2nd case: E = q2 + w2

E in both cases is the same , it follows that q1≠ q2

heat from this simple example, is associated with a given process, depends on how the process is carried out.

We cannot write q = qf- qi

In summary, heat and work are not state functions because they are not properties of a system.

E is a state function although q and w are not state functions.

Example 6.2

Enthalpy of Chemical Reactions:

We will consider two situations

1- If a chemical reaction is run at constant volume: then V = 0 and no work will result from this change.

E = q - PV = qv

Where qv is the heat of reaction at constant volume. qv = E

2- If a chemical reaction is run at constant pressure:

E = q + w

E = qp - PV

qp = E + PV

Whereqp is the heat of reaction at constant pressure.

Enthalpy

Enthalpy is defined as the system’s internal energyplus the product of its pressure and volume.

H = E + PV

E,P and V are all state functions, that is, the change in (E + PV) depend on the initial and final states. The change in H or H depends on the initial and final states. Thus H is a state function.

H = E + (PV)

H = E + PV

We see that for a constant-pressure process, qp = H

Enthalpy of Reactions:

Most of the reactions are carried out at constant pressure, and for any reaction of the type

Reactants  Products

We define the change in enthalpy, called

Enthalpy of the reaction,H: is the difference between the enthalpies of the products and the enthalpies of the reactants.

H = H (products) – H (reactants)

  • For endothermic: H > 0 heat absorbed from the surroundings by the system.
  • For exothermic: H < 0 heat released by the system from the surroundings.

Thermochemical Equations: stoichiometric equations with reaction enthalpy

At 0oC and 1 atm pressure ice melts to liquid

H2O(s)  H2O(l)H = 6.01 kJ

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)H = - 890.4 kJ

If you reverse a reaction, the sign of H changes:

H2O(l)  H2O(s)H= - 6.01 kJ

CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)H = 890.4 kJ

If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O(s)  2H2O(l)H = 2 x 6.01 kJ

The physical states of all reactants and products must be specified in thermochemical equations.

H2O(s)  H2O(l)H = 6.01 kJ

H2O(l)  H2O(g)H = 44.0 kJ

CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)H = - 802.4 kJ

Example 6.3

A comparison of H and E:

Let us consider the following reaction:

2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)H = - 367.5 kJ

To calculate the change in internal energy, we rearrange equation as:

E = H - PV

assume the temperature to be 25oC, we can show that the volume of 1 mole of H2 gas at 1 atm is 24.5 L, so that - PV = 24.5 L.atm or - 2.5 kJ

E = - 367.5 kJ - 2.5 kJ = - 370.0 kJ

The calculation shows that E and H are approximately the same. The reason H is smaller than E is that some of internal energy released is used to do gas expansion work, so less heat is evolved. For the reactions that do not involve gases, Vis usually very small and so E is practically the same as H.

Another way to calculate the internal energy change of a gaseous reaction is to assume ideal gas behavior and constant temperature. In this case,

E = H - (PV)

E = H - (nRT)

E = H - RTn

n = number of moles of product gases - number of moles of reactant gases

Example 6.4

Calorimetry: the measurements of heat changes.

A Calorimeter is an instrument used to measure the changes in physical and chemical process

Heat changes depend on specific heat and heat capacity.

The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius.

The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius.

C = ms where m is the mass of the substance in grams

Specific heat is an intensive property, whereas heat capacity is an extensive property.

Heat change (q)= mstq= Ct

t = tf - ti

Example 6.5

Constant-Volume Calorimetry:

A bomb calorimeter: is used to measure the energy released in exothermic reactions at constant volume. It gives very precise measurements.

The calorimeter is filled with oxygen gas before it is placed in the bucket.

The sample is ignited electrically, and the heat produced by the reaction can be accurately determined by measuring the temperature increase in the known amount of surrounding water.

No heat or mass is lost to the surroundings during the time it takes to make measurements. Therefore we can call the bomb and the water (in which it is submerged) an isolated system.

Because no heat enters or leaves the system throughout the process, the heat change of the system (qsystem) must be zero.

qsystem = qcal + qrxn = 0

where qcal and qrxn are the heat changes of the calorimeter and the reaction, respectively. Thus

qrxn = - qcal

qcal = Ccalt

Example 6.6

Constant-Pressure Calorimetry:

A coffee cup calorimeter: is made of two Styrofoam coffee cups.

The outer cup helps to insulate the reacting mixture from the surroundings.

Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter.

The heat produced or absorbed by the reaction can be determined by measuring the temperature change.

Assume that no heat is lost to the surroundings (treat the calorimeter as an isolated system) and ignore the small heat capacity of the calorimeter, we write

qsystem = qsoln + qrxn = 0

qrxn = - qsoln

Example 6.7

No way to measure the absolute value of the enthalpy of a substance. Chemists establish the reference point for all enthalpy expressions which is called, standard enthalpy of formation (Hfo).

Standard enthalpy of formation (Hfo) is the heat change that results when one mole of a compound is formed from its elementsat a pressure of 1 atm.

The standard enthalpy of formation of any element in its most stable form is zero.

Hfo(O2) = 0Hfo(C, graphite) = 0

Hfo(O3) ≠ 0Hfo(C, diamond) ≠ 0

Table 6.4 lists the standard enthalpies of formation for a number of elements and compounds.

The standard enthalpy of reaction(Hrxno) is the enthalpy of a reaction carried out at 1 atm.

aA + bB  cC + dD

Hrxno = [cHfo(C) + dHfo(D)] - [aHfo(A) + bHfo(B)]

In general:

Hrxno = Σ nHfo(products) - Σ mHfo(reactants)

The direct method:

C(graphite) + O2(g)  CO2(g)Hrxno = - 393.5kJ

Hrxno = (1 mol)Hfo(CO2, g) – [(1 mol)Hfo(C, graphite) + (1 mol)Hfo(O2, g)]

Hrxno = (1 mol)Hfo(CO2, g) = - 393.5kJ

Hfo(CO2, g) = - 393.5kJ

Other compounds that can be studied by direct method are SF6, P4O10, and CS2.

The Indirect Method:

Hfo can be determined by an indirect approach, which is based on the law of heat summation (or simply Hess’s law).

Hess’s Law:When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

Hess’s Lawis based on the fact that because H (Enthalpy) is a state function, Hdepends only on the initial and final state.

Examples 6.8, 6.9.

Selected Problems: 14, 22, 32, 36, 44, 48, 49, 51, 52, 54, 56, 58, 61, 64, 74, 92.

1