Thermochemistry © 2009, Prentice-Hall, Inc. Chapter 5 Thermochemistry Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay,

Thermochemistry

© 2009, Prentice-Hall, Inc.

Chapter 5Thermochemistry

Chemistry, The Central Science, 11th editionTheodore L. Brown; H. Eugene LeMay, Jr.;

and Bruce E. Bursten

Thermochemistry


Overview of Thermochemistry

• System vs. Surroundings

• Energy vs. Work

• Potential energy vs. Kinetic Energy

• Changes in internal energy endothermic vs. exothermic changes

Thermochemistry

Energy and Chemical Reactions

• Energy changes are due to rearranging chemical bonds.

• Energy of chemical bonding = a form of chemical potential energy: breaking of existing bonds and forming of new bonds.

• Energy is always needed to break bonds. Energy release occurs when new bonds are formed.

• The balance between these two processes overall reaction is endothermic or exothermic.


Thermochemistry


Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: the calorie (cal).

1 cal = 4.184 J

1 J = 1 kg m2

s2

Thermochemistry


E, q, w, and Their Signs

Thermochemistry

Work

• w = F x d

• In chemical systems, only P/V work (contraction or expansion of a gas) is of significance, and only when there is an increase or a decrease in the amount of gas present. There must be a change in volume in order for the system to do work on the surroundings (or have work done on it by the surroundings).


Thermochemistry


State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

Thermochemistry


State Functions• However, we do know that the internal energy

of a system is independent of the path by which the system achieved that state.– In the system below, the water could have reached

room temperature from either direction.

Thermochemistry


State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the

system, not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

Thermochemistry


Enthalpy

• Enthalpy = the heat transferred between the system and the surroundings during a chemical reaction carried out under constant pressure

= is the internal energy plus the product of pressure and volume:

H = E + PV

Thermochemistry


Enthalpy

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (E + PV)

• This can be written

H = E + PV

Thermochemistry


Enthalpy

• Since E = q + w and w = -PV, we can substitute these into the enthalpy expression:

H = E + PV

H = (q+w) − w

H = q

• So, at constant pressure, the change in enthalpy is the heat gained or lost.

Thermochemistry


Enthalpy

• Heat from surroundings system

endothermic, H > 0

• Heat from system surroundings

exothermic, H < 0

Enthalpy is a state function – doesn’t matter how it got there.

Thermochemistry


Endothermicity and Exothermicity

• A process is endothermic when H is positive.

Thermochemistry


Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Thermochemistry


Definitions:System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

Thermochemistry


Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

Thermochemistry


Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

• When heat is released by the system into the surroundings, the process is exothermic.

Thermochemistry


Enthalpy

H = change in enthalpy

= the heat transferred between the system and surroundings during a chemical reaction carried out under constant pressure

Thermochemistry


Enthalpy

• Heat from surroundings system

endothermic, H > 0

• Heat from system surroundings

exothermic, H < 0

Enthalpy is a state function – doesn’t matter how it got there.

Thermochemistry


Overview of Thermochemistry

Our focus:

• Enthalpy is a state function.

• H and stoichiometry

• Specific Heat q = c x m x T (and related calculations, including calorimetry)

• Hess’s Law and its application, including enthalpies of formation

Chapter 19: Entropy and Gibbs Free Energy (second semester)

Thermochemistry


Enthalpy of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Thermochemistry


Enthalpy of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

Thermochemistry


The Truth about Enthalpy

1. Enthalpy is an extensive property – the magnitude of enthalpy is directly proportional to the amount of reactant consumed.

Thermochemistry



2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction.

Example:

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H = –890 kJ,

But CO2(g) + 2H2O(l) CH4(g) + 2O2(g)

H = +890 kJ.

Thermochemistry



3. H for a reaction depends on the state of the products and the state of the reactants.

2H2O(g) 2H2O(l) H = –88 kJ

Thermochemistry


Sample Exercise 5.4 (p. 179)

How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system?

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

H = –890 kJ

(-250 kJ)

Thermochemistry


Practice Exercise 5.4

Hydrogen peroxide can decompose to water and oxygen by the following reaction:

2 H2O2(l) 2 H2O(l) + O2(g) H = -196 kJ

Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.

(-14.4 kJ)

Thermochemistry


Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Thermochemistry


Heat Capacity and Specific Heat

The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

Thermochemistry


Heat Capacity and Specific HeatMolar heat capacity = the heat capacity of 1 mol of a substance

Specific heat capacity (or simply specific heat) = the heat capacity of 1 g of a substance.

Thermochemistry


Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

c =q

m T

Thermochemistry


• Another useful version of this equation:

q = (specific heat) x (grams of substance) x T

= c x m x T

Thermochemistry


Sample Problem 5.5 (p. 181)

a) How much heat is needed to warm 250 g of water from 22oC to near its boiling point, 98oC? The specific heat of water is 4.18 J/g-K.

(7.9 x 104 J)

b) What is the molar heat capacity of water?

(75.2 J/mol-K)

Thermochemistry


Practice Problem 5.5a) Large beds of rocks are used in some solar-

heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0oC.(4.9 x 105 J)

b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?(11k = 11oC decrease)

Thermochemistry


Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

Thermochemistry



Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation:

qsoln = m c T = -qrxn

Note the reversal of the sign of q.

Thermochemistry



For dilute aqueous solutions, the specific heat of the solution will be close to that of pure water.

Thermochemistry


Sample Exercise 5.6 (p.182)When a student mixes 50. mL of 1.0 M HCl and 50. mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0oC to 27.5oC. Calculate the enthalpy change for the reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100. mL, that its density is 1.00 g/mL, and that its specific heat is 4.18 J/g-K.

(-54 kJ/mol)

Thermochemistry


Practice Exercise 5.6When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by the following reaction:

AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq)

Calculate H for this reaction, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g-oC.(-68 kJ/mol)

Thermochemistry


Bomb Calorimetry

• Reactions can be carried out in a sealed “bomb” such as this one.

• The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.

Thermochemistry


Bomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

Thermochemistry



Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of methylhydrazine with oxygen produces N2(g), CO2(g), and H2O(l):

2 CH6N2(l) + 5 O2(g) 2 N2(g) + 2 CO2(g) + 6 H2O(l)

When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00oC to 39.50oC. In a separate experiment the heat capacity of the bomb calorimeter is measured to be 7.794 kJ/oC. What is the heat of reaction for the combustion of a mole of CH6N2 in this calorimeter?

(-1.30 x 103 kJ/mol CH6N2)

Thermochemistry



A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/oC. The temperature increases from 23.10oC to 24.95oC. Calculate the heat of combustion of

a) lactic acid per gram (-15.2 kJ/g) and

b) per mole (-1370 kJ/mol)

Thermochemistry


Hess’s Law

H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.

• However, we can estimate H using published H values and the properties of enthalpy.

Thermochemistry


Hess’s Law

Hess’s law states that “if a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”

Thermochemistry


Hess’s Law

Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products.

Thermochemistry


Hess’s Law

Combustion of methane

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = –802 kJ

2H2O(g) 2H2O(l) H = –88 kJ__________________________________________

CH4(g) + 2O2(g) CO2(g) + 2H2O(l) H = –890 kJ

Thermochemistry



The enthalpy of combustion of C to CO2 is -393.5 kJ/mol C, and the enthalpy of combustion of CO to CO2 is -283.0 kJ/mol CO:

1) C(s) + O2(g) CO2(g) H1 = -393.5 kJ2) CO(g) + ½ O2(g) CO2(g) H2 = -283.0 kJ

Using these data, calculate the enthalpy of combustion of C to CO:

3) C(s) + ½ O2(g) CO(g) H3 = ?

(-110.5 kJ)

Thermochemistry


Practice Exercise 5.8HW

Carbon occurs in two forms, graphite and diamond. The enthalpy of combustion of graphite is -393.4 kJ/mol and that of diamond is -395.4 kJ:

C(graphite) + O2(g) CO2(g) H1 = -393.5 kJ

C(diamond) + O2(g) CO2(g) H2 = -395.4 kJ

Calculate H for the conversion of graphite to diamond:

C(graphite) C(diamond) H3 = ?

(+1.9 kJ)

Thermochemistry


Sample Exercise 5.9 (p. 187)HW

Calculate H for the reaction 2 C(s) + H2(g) C2H2(g)

Given the following reactions and their respective enthalpy changes:

C2H2(g) + 5/2 O2(g) 2 CO2(g) + H2O(l) H = -1299.6 kJ

C(s) + O2(g) CO2(g) H = -393.5 kJH2(g) + ½ O2(g) H2O(l) H = -285.8

kJ

(226.8 kJ)

Thermochemistry



Calculate H for the reactionNO(g) + O(g) NO2(g)

given the following reactions and their respective enthalpy changes:

NO(g) + O3(g) NO2(g) + O2(g) H = -198.9 kJO3(g) 3/2 O2(g) H = -142.3 kJO2(g) 2 O(g) H = 495.0

kJ

(-304.1 kJ)

Thermochemistry


Enthalpies of Formation

An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.

Thermochemistry


Standard Enthalpies of Formation

Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure).

Thermochemistry

Heats of Formation

• We can use the heats of formation to predict the H of a particular reaction – another version of Hess’s Law


Thermochemistry



For which of the following reactions at 25oC would the enthalpy change represent a standard enthalpy of formation? For those where it does not, what changes would need to be made in the reaction conditions?

a) 2 Na(s) + ½ O2(g) Na2O(s)

b) 2 K(l) + Cl2(g) 2 KCl(s)

c) C6H12O6(s) 6 C(diamond) + 6 H2(g) + 3 O2(g)

Thermochemistry



Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4).

Thermochemistry


Calculation of H

• Imagine this as occurringin three steps:

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

C3H8 (g) 3 C (graphite) + 4 H2 (g)

3 C (graphite) + 3 O2 (g) 3 CO2 (g)

4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


Calculation of H


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


Calculation of H


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

Thermochemistry


C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)



4 H2 (g) + 2 O2 (g) 4 H2O (l)

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

• The sum of these equations is:

Thermochemistry


Calculation of H

We can use Hess’s law in this way:

H = nHf°products – mHf° reactants

where n and m are the stoichiometric coefficients.

Thermochemistry


H = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]

= [(-1180.5 kJ) + (-1143.2 kJ)] – [(-103.85 kJ) + (0 kJ)]= (-2323.7 kJ) – (-103.85 kJ) = -2219.9 kJ

C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)

Calculation of H

Thermochemistry



a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l).(-3267 kJ)

b) Compare the quantity of heat produced by combustion of 1.00 g propane (C3H8) to that produced by 1.00 g benzene.

(C3H8(g): -50.3 kJ/g; C6H6(l): -41.8 kJ/g)

Thermochemistry



Using the standard enthalpies of formation listed in Table 5.3 or Appendix C, calculate the enthalpy change for the combustion of 1 mol of ethanol:

C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(l)

(-1367kJ)

Thermochemistry



The standard enthalpy change for the reaction

CaCO3(s) CaO(s) + CO2(g)

is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO2(g) given in Table 5.3, calculate the standard enthalpy of formation of CaCO3(s).

(-1207.1 kJ/mol)

Thermochemistry



Given the following standard enthalpy of reaction, use the standard enthalpies of formation in Table 5.3/Appendix C to calculate the standard enthalpy of formation of CuO(s):

CuO(s) + H2(g) Cu(s) + H2O(l) Ho = -129.7 kJ

(-156.1 kJ/mol)

Thermochemistry


Sample Integrative Exercise (p. 186)

Trinitroglycerin, C3H5N3O9, (usually referred to simply as nitroglycerin) has been widely used as an explosive. Alfred Nobel used it to make dynamite in 1866. Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart), by dilating the blood vessels. The enthalpy of decomposition at 1 atmosphere pressure of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas at 25oC is -1541.4 kJ/mol.

Thermochemistry



a)Write a balanced chemical equation for the decomposition of trinitroglycerin.

b)Calculate the standard heat of formation of trinitroglycerin.

c)A standard dose of trinitroglycerin for relief of angina is 0.60 mg. Assuming that the sample is eventually completely combusted in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released?

Thermochemistry



d)One common form of trinitroglycerin melts at about 3oC. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain.

e)Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to break rockfaces in highway construction.

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Thermochemistry © 2009, Prentice-Hall, Inc. Chapter 5 Thermochemistry Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay,