Matter and Energy
The universe is composed of matter and energy. Matter includes all tangible things, and has mass and volume which can be measured. The concept of energy is more difficult to grasp because energy is intangible. Energy, unlike matter, cannot be held in your hand.

Energy can be defined as the capacity to do work (move matter) or produce heat. A wound clock acquires "something" with which it can do work. This "something" that enables the clock to do work is energy. An object can exhibit energy in two fundamental ways, kinetic energy (E_k) and potential energy (E_P).

Kinetic energy is the energy of motion an object associated with mechanical work, and is described mathematically
by the equation; E_K = ½ mv², where m is mass and v is velocity.

Potential energy is stored energy, it is energy related to position. An object has potential energy by virtue of its position in a field of force. A 1 kg object held 1 m above the surface of the earth has a potential energy of E_P = mgh = (1 kg)(9.8 m/s²)(1 m) = 1 kg×m²/s². Potential energy can be thought of as work already done.

The SI unit of energy is a derived unit called the joule (J). The SI natural units of energy are kg×m²/s². (1 J = 1 kg×m²/s² )
1 calorie (cal) = 4.184 J

Example 1
How many joules are in 8.32 cal?     [34.8 J]

Work is defined by the mathematical relationship: work = force x distance. The SI unit of force is the newton (N). 1N = 1 kg×m/s². Energy and work have the same units. work = force x distance = 1 N x 1 m = 1 kg×m²/s² = 1 J

Chemists define work as directed energy change resulting from a process. Chemical processes (reactions) are almost always accompanied with the absorption or release of one form of energy, heat (thermal energy). The study of the energy (heat) change associated with chemical reactions is known as thermochemistry.

Temperature is a relative measure of how hot or how cold an object is. It is a measure of the average random motion (kinetic energy) of the unit particles of an object. It is the property of an object that determines the direction in which thermal energy will be transferred when it is in contact with another object at a different temperature. When two objects are in contact with one another, and at the same temperature, the average kinetic energy of the unit particles of the two objects is equal.

Heat (q) is the thermal energy that "flows" into or out of a substance due to a temperature difference. Heat flows spontaneously from the warmer object to the colder object.

The internal energy (E) of a system is the sum of the kinetic energy (E_k) and potential energy (E_P) of all the unit particles (atoms, molecules, ions) of the system. E  = E_k+ E_P The following table will perhaps aid in understanding the concept and origin of the energies of a system’s composition.
 

The kinetic energy (thermal energy) is associated with random molecular motion. There are three types.	The potential energy (chemical energy) is associated with electrostatic attractions within and between molecules. There are two types.
1. Tranlational	1. Intramolecular forces (bonds)
2. Rotational	2. Intermolecular forces
3. Vibrational

The Greek letter D (delta) is used to indicate changes in state functions. Thus,  DE = E_final - E_initial  is the change of internal energy between initial and final states.

A positive value for DE (DE > 0) means internal energy increases.
Energy is added to the system.

A negative value for DE (DE < 0) means internal energy decreases.
Energy leaves the system.

Stated mathematically in terms of internal energy, heat and work, the law of conservation of energy is referred to as the first law of thermodynamics and is expressed as: DE_system = q_system + w_system

In this equation, q_system is the quantity of energy transferred by heating the system and  w_system is quantity of energy transferred by doing work on the system.

These two thermodynamic quantities will have a magnitude and a sign. The sign conventions are as follows:
    If heat is transferred into the system from the surroundings, then q is assigned a positive sign, +q.
    If heat is transferred out of the system to the surroundings, then q is assigned a negative sign, -q.
    If work is done on the system by the surroundings, then w is assigned a positive sign, +w.
   If work is done by the system on the surroundings, then w is assigned a negative sign, -w.

Example 2
A gas absorbs 35.0 J of heat and does 15.0 J of work. What is DE?

DE = q + w = (+35.0 J) + (-15.0 J) = +20.0 J

Example 3
How much heat is required to raise the temperature of a 850 gram block of aluminum from 22.8^oC to 94.6^oC?
The specific heat of Al is 0.902 J/g^.oC.
    q = (mass)(specific heat)(DT) = (850 g)(0.902 J/g^.oC)(94.6^oC-22.8^oC) = + 55.0 kJ

Example 4
What is the specific heat of iron at 25^oC if 285 J of heat were transferred when a 33.69 gram sample of iron cooled from
43.8^o0C to 25^oC?
    q = (mass)(specific heat)(DT)
    -285 J = (33.69 g)(specific heat)(25.0^oC - 43.80^oC)
    specific heat = 0.45 J/g^.oC

Example 5
A 25.88 g sample of a metal was heated to 85.32 ^oC and then dropped into 35.14 g of water at 22.48 ^oC. The temperature of the water rose to 26.47 ^oC. What is the identity of the metal?  Al 0.902 J/g^.oC;   Cu 0.385 24.5 J/g^.oC;   Fe 0.449 J/g^.oC;   Pb 0.128 J/g^.oC;   Ag 0.235 J/g^.oC
[Answer pecific Heat = 0.385 J/g^.oC     The metal is copper.]

Example 6
Calculate the final temperature when 25.00 g of water at 20.0 ^oC is mixed with 75.00 g of water at 40.00 ^oC.
[T_Final = 35.0 ^oC]

In any phase transition, heat (q) will be transferred as the substance undergoes the transition.
    Vaporization is an endothermic process.  Endothermic processes occur when heat energy is transferred into a system.

        H₂O(l)  ---> H₂O(g) q = +40.7 kJ/mol

    Condensation is an exothermic process.   Exothermic processes occur when heat energy is transferred out of a system.

        H₂O(g)  ---> H₂O(l) q = -40.7 kJ/mol

Example 7
How much heat is required to vaporize 25.0 g of carbon disulfide at 25^oC?
The heat of vaporization (DH_vap) for carbon disulfide is +27.4 kJ/mol.
q_vap =  (mole)(DH_vap)  = (25.0 g)(1mol/76.15 g)(27.4 kJ/mol)  = 9.00 kJ
 

Example 8
Liquid butane, C₄H₁₀, is stored in cylinders to be used as a fuel.  Suppose 31.4 g of butane gas is removed from a cylinder.  How much heat must be provided to vaporize this much gas?  The heat of vaporization of butane is 21.3 kJ/mol.
q_vap =  (mole)(DH_vap)  = (31.4 g)(1 mol/ 58.14 g)(21.3 kJ/mol)  = 11.5 kJ

A negative value (-) for DH indicates an exothermic reaction, that is heat is produced by the reaction system.
    CH₄(g) + 2 O₂(g) ---> CO₂(g) + 2 H₂0(g) DH = - 890 kJ

A positive value (+) for DH indicates an endothermic reaction, that is heat is absorbed by the reaction system.
    CH₃OH(l) ---> CO(g) + 2H₂(g)  DH = +90.7 kJ

The value of DH depends upon the phase of the reactants and the products.
    2H₂ (g) + O₂ (g) ---> 2H₂O(g) DH = -483.7 kJ
    2H₂(g) + O₂(g) ----> 2H₂O(l) DH = -571.7 kJ

Stoichometric Calculations Involving Thermochemical Equations
    1.    When a thermochemical equation is multiplied by any factor, the value of DH for the new equation is obtained by
            multiplying the value of DH in the original equation by that same factor.
    2.     When a chemical equation if reversed, the value of DH is reversed in sign.

EXAMPLE 3
Using the following thermochemical equation, calculate how much heat is associated with the decomposition of
4.00 moles of NH₄Cl.

NH₃(g) + HCl(g) ---> NH₄Cl(s)    DH = - 176 kJ

(4.00 mol NH₄Cl)(+176 kJ/1 mol NH₄Cl) = +704 kJ

EXAMPLE 4
How much heat is associated with the synthesis of 45.0 g of NH₃ according to the following equation?
4 NO(g) + 6 H₂O(l) ---> 4 NH₃(g) + 5 O₂(g)    DH_rxn = +1170 kJ

(45.0 g NH₃)(1 mol/17.04 g NH₃) = 2.641 mol NH₃

(2.641 mol NH₃)(+1170kJ/4 mol NH₃) = +772 kJ

EXAMPLE 5
Calculate the mass of ethane, C₂H₆, which must be burned to produce 100 kJ of heat.
2 C₂H₆(g) + 7 O₂(g) ---> 4 CO₂(g) + 6 H₂O(l)   DH = - 3120 kJ

(-100 kJ)(2 mol C₂H₆ / -3120 kJ) = 0.0641 mol C₂H₆

(0.0641 mol C₂H₆)(30.08 g C₂H₆/1 mol C₂H₆) = 1.93 g C₂H₆

Bomb Calorimetry: Reactions at Constant Volume
The amount of heat given off by the combustion of a fuel can be determined very accurately in the so-called bomb calorimeter, which consists of a combustion chamber (the "bomb") set in another chamber filled with water. Heat generated by combustion of the fuel is transmitted to the water, raising its temperature. The calorie content of food is tested this way.

Bomb calorimeters are usually composed of two parts, the bomb itself wherein the reaction takes place and the water which surrounds the bomb. The calculations for determining the heat of reaction require that the heat change of the bomb and the water be considered individually.

    q_calorimeter = heat capacity of the calorimeter x DT

    q_system = q_calorimeter +  q_rxn

    Because no heat enters or leaves the system q_system = 0

    q_rxn = -q_calorimeter=  - C_calorimeter(DT)

or another way of looking at it is that the heat associated with the reaction will be transferred to the calorimeter so that the heat will have the same magnitude but opposite sign.

The amount of heat involved in a reaction can be determined from:

    q_rxn = q_water +  q_bomb

    q_water = measured heat change of the water.

    q_bomb = measured heat change of bomb.

    Thus, q_water and q_bomb must be formed before q_rxn can be determined.

    q_water = (specific heat of water)(mass of water)(DT)

    q_calorimeter = (heat capacity of the bomb)(DT)

EXAMPLE 11
Exactly 3.00 g of graphite is burned to CO₂ in a copper calorimeter. The initial temp is 25.0^oC and the final temp was 36.3^oC. The heat capacity of the calorimeter is 8.887 kJ/^oC.  What is the thermochemical equation for the combustion of graphite?

    A.    Calculate the heat transferred to the calorimeter.
            q_calorimeter = C(DT) = (8.887 kJ/^oC)(36.3^oC - 25.0^oC) = +100.4 kJ

            The energy change for the reaction is equal in magnitude and opposite in sign to the heat energy required to raise the
            temperature of the calorimeter. This means that q_rxnis -100.4 kJ.

    B.    Derive the thermochemical equation for the combustion of graphite.
            C(s) + O₂(g) ---> CO₂(g)  DH = ?

            (-100.4 kJ)/[(3.00 g C)(1 mol C/12.01 g C)] = -401.9 kJ/mol C

            (1 mole C(s))(-401.9 kJ/mol C) = -401.9 kJ

            C(s) + O₂(g) ---> CO₂(g)  DH = -401.9 kJ

EXAMPLE 12
A certain calorimeter has a heat capacity of 16.35 kJ/^oC. What is the final temperature of the calorimeter if 17.038 g of kerosene (C₁₂H₂₆) is burned in the calorimeter? The initial temperature of the calorimeter is 25.00 ^oC. The heat of combustion of kerosene is shown in the equation below.

    C₁₂H₂₆(l) + 37/2 O₂(g) ---> 12 CO₂(g) + 13 H₂O(l)  DH = -7513 kJ

    A.     (17.038g C₁₂H₂₆)(1 molC₁₂H₂₆/170.38g C₁₂H₂₆)(-7513 kJ/mol C₁₂H₂₆) = -751.3 kJ

    B.     q_rxn = -q_cal = -C(DT)

            -751.3 kJ = -(16.35 kJ/^oC)(T_f - 25.00^oC)

            T_f  =  70.95^oC

Hess's law states that the heat transferred in a given change is the same whether the change takes place in a single step or in several steps.

DH_rxn for a given reaction is the same whether that reaction takes place directly through one step or via several different reactions.

    S(s) + O₂(g) ---> SO₂(g) DH = -296 kJ

    SO₂(g)  + 1/2 O₂(g) ---> SO₃(g)   DH = -98.9 kJ
    ____________________________________________
    S(s) + 1 1/2 O₂(g)  ---> SO₃(g)    DH = -394.9 kJ

In order to use Hess's Law:

1) If a rxn is reversed, the sign of  DH is reversed.

    S(s) + O₂(g) ---> SO₂(g) DH = -296 kJ

    SO₂(g)  ---> S(s)  +  O₂(g)   DH = +296 kJ

2) If the coefficients in a balanced equation are multiplied by some number, the value of  DH must be multiplied by that same number.

    S(s) + O₂(g)  ---> SO₂(g)    DH = -296 kJ

    2 S(s) + 2 O₂(g)  ---> 2 SO₂(g)     DH = (2)(-296 kJ) = -592 kJ

EXAMPLE 13
Calculate DH for the vaporization of water from liquid using the thermochemical equations given.

H₂O(l) ---> H₂O(g) DH = ?

    a) H₂(g)  + 1/2 O₂(g)   --->H₂O(l)  DH = -286 kJ

    b) H₂(g)  + 1/2 O₂(g)   ---> H₂O(g)   DH = -242 kJ
 

            H₂O(l) ---> H₂(g)  + 1/2 O₂(g)  DH = +286 kJ

            H₂(g)  + 1/2 O₂(g)   ---> H₂O(g) DH = -242 kJ
        _____________________________________
            H₂O(l)  ---->  H₂O(g) DH = +44 kJ

EXAMPLE 14
Calculate DH for the following reaction using the thermochemical equations given.

2 C_(s) + H_{2 (g)} ---> C₂H_{2 (g)}  DH = ?

    a) C_(s) + O_{2 (g)}  ---> CO_{2 (g)}  DH = -393.5 kJ

    b) H_{2 (g)} + 1/2 O_{2 (g)}  ---> H₂O_(l)  DH = -285.8 k J

    c) 2 C₂H_{2 (g)} + 5 O_{2 (g)}  ---> 4 CO_{2 (g)} + 2 H₂O_(l) DH = -2598.8 kJ

    Solution:

                2 C_(s) + 2 O_{2 (g)}  ---> 2 CO_{2 (g)}DH = -787.0 kJ

                H_{2 (g)} + 1/2 O_{2 (g)}  ---> H₂O_(l)DH = -285.8 k J

                2 CO_{2 (g)} + H₂O_(l) ---> C₂H_{2 (g)} + 2 1/2 O_{2 (g)}  DH = +1299.4 kJ
    _____________________________________________________
                2 C_(s) + H_{2 (g)}  ---> C₂H_{2 (g)}DH = +226.6 kJ

For a gas, the standard state is a pressure of exactly 1 atmosphere.
For a substance present in solution, the standard state is a concentration of exactly 1 M.
For a pure substance in a condensed state (liquid or solid), the standard state is the pure liquid or solid.
For an element, the standard state is the form in which the element exists under conditions of 1 atm and 25^oC.

The standard enthalpies of formation of the elements in their standard states are zero.

The standard enthalpy change, DH^o, for any chemical reaction is found by subtracting the sum of the heats of formation of the reactants from the sum of the heats of formation of the products.  This is shown in the following equation.

DH^o= SDH^o_f (Products) - SDH^o_f (Reactants)                [NOTE:  "S" is the symbol which means "sum".]

EXAMPLE 15
Using the standard heats of formation, calculate the standard heat of reaction, DH^o , for the following reaction.

    2 NO_{2 (g)} ---> N₂O_{4 (g)} DH^o  = ?

DH^o_f    NO_{2 (g)}= +33.9 kJ/mol;       DH^o_f     N₂O_{4 (g)} = +9.7 kJ/mol

DH^o= SDH^o_f (Products) - SDH^o_f (Reactants)

DH^o = [(1 mol N₂O₄)(9.7 kJ/mol)] - [(2 mol NO₂)(33.9 kJ/mol)] = -58.1 kJ

EXAMPLE 16
Using the standard heats of formation, calculate the standard heat of reaction, DH^o , for the following reaction.

C₇H_{16 (l)}+ 11 O_{2 (g)} ----> 7 CO_{2 (g)} + 8 H₂O_(l)   DH^o  = ?

DH^o_f  C₇H_{16 (l)} = -198.8 kJ/mol;    DH^o_f  CO_{2 (g)} = -393.5 kJ/mol;       DH^o_f H₂O_(l) = -285.9 kJ/mol

DH^o= SDH^o_f (Products) - SDH^o_f (Reactants)

DH^o  = [(7 mol CO₂)(-393.5 kJ/mol) + (8 mol H₂O_(l))(-285.9 kJ/mol)]  -  [(1 mol C₇H_{16 (l)})(-198.8 kJ/mol)]

                        = -5041.7 kJ - (-198.8 kJ) = -4842.9 kJ
 

CHEMISTRY 101 THERMOCHEMISTRY SAMPLE PROBLEMS

1. How much heat is liberated when 6.00 mol of Fe₂O₃(s) are reacted according to the following reaction? (-21.4 kcal)
3 Fe₂O₃(s) + CO(g) --> 2 Fe₃O₄(s) + CO₂(g)  DH = -10.7 kcal

2. What amount of heat is associated with the reaction of 43.77 g of NaCl with sulfuric acid? (+23.81 kJ)
H₂SO₄(l) + 2 NaCl(s) --> Na₂SO₄(s) + 2 HCl(g)  DH = +63.60 kJ

3. A 466 g sample of water is heated from 8.50C to 74.60C. Calculate the amount of heat absorbed by the water. The specific heat of water is 4.184 g^oC. (129 kJ)

4. An electrical heater is used to supply 25.0 J of energy to a 25.0 g sample of Ag originally at 22^oC. Calculate the final temperature. The specific heat of Ag is 0.2349 g^oC. (26^oC)

5. Assuming no heat loss to the surrounding or to the container, calculate the final temperature when 100 g of silver at 40.0^oC is immersed in 60.0 g of water at 10.0^oC. The specific heat of silver is 0.2349 J/g^oC and for water is 4.184 J/g^oC. (12.6^oC)

6. Determine DH^ofor the following reaction of burning ethyl alcohol in oxygen:

C₂H₅OH(l) + 3 O₂(g) ---> 2 CO₂(g) + 3 H₂O (l)   DH = ? (-327.2 kcal)

DH_f of C₂H₅OH(l) = -65.9 kcal/mol;   DH_f of CO₂(g) = -94.1 kcal/mol;  DH_f of H₂O(l) = -68.3 kcal/mol

7. From the following equations and the enthalpy changes, calculate the enthalpy of reaction of iron(III) oxide, Fe₂O₃, with carbon monoxide:

Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)   DH = ? (-26.7 kJ)

(a) CO(g) + 1/2 O₂(g) ---> CO₂(g) DH = -283.0 kJ

(b) 2 Fe(s) + 3/2 O₂(g) ---> Fe₂O₃(s) DH = -822.3 kJ

8. Using the thermochemical equations given, calculate the heat of hydrogenation of acetylene.
C₂H₂(g) + 2 H₂(g) --> C₂H₆(g)   DH = ? (-312 kJ)

(a) 2 C₂H₂(g) + 5 O₂(g) --> 4 CO₂ + 2 H₂O(g)   DH = -2602 kJ

(b) 2 C₂H₆(g) + 7 O₂(g) --> 4 CO₂ + 6 H₂O(g)   DH = -3123 kJ

(c) 2 H₂(g) + O₂(g) --> 2 H₂O(g) DH = -572 kJ

9. The combustion of 1 mol of benzene, C₆H₆(l), to produce CO₂(g) and H₂O(l) liberates 3271 kJ. Given the the DH_f of CO₂(g) and H₂O(l) are -394 kJ/mol and -286 kJ/mol respectively, calculate the heat of formation of benzene. (+49 kJ/mol)

CALORIMETRY
10. A quantity of 1.435 g of naphthalene (C₁₀H₈) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.17C to 25.84C. If the quantity of water surrounding the calorimeter was exactly 2000 g and the heat capacity of the bomb calorimeter was 1.80 kJ/C, calculate the heat of combustion of naphthalene on a molar basis; that is , find the molar heat of combustion. (-5.15 x 10³ kJ/mol)

11. A quantity of 1.00 x 10² mL of 0.500 M HCl is mixed with 1.00 x 10² mL of 0.500 M NaOH in a constant-pressure calorimeter having a heat capacity of 335 J/C. The initial temperature of the HCl and NaOH solutions is the same, 22.50C, and the final temperature of the mixed solution is 24.90C. Calculate the heat change for the neutralization reaction
NaOH(aq) + HCl(aq) ---> NaCl(aq) + H₂O(l)

Assume that the densities and specific heats of the solutions are the same as for water (1.00g/mL and 4.182 J/gC, respectively). (-56.2 kJ/mol)

WORK
12. A certain gas initially at room temperature undergoes an expansion in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against vacuum and (b) against a constant pressure of 1.2 atm.     1 L^.atm = 101.325 J              ((a) 0 (b) -4.8 Latm = -4.9 x 10² J)

13. The work done when a gas is compressed in a cylinder is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process. (334 J)

14. The oxidation of nitric oxide to nitrogen dioxide is a key step in the formation of smog:     2NO(g) + O₂(g) ---> 2NO₂(g)  DH = -113.1 kJ
If 6.00 moles of NO react with 3.00 moles of O₂ at 1.00 atm and 25C to form NO₂, calculate the work done (in kilojoules) against a pressure of 1.00 atm. What is the DU for the reaction? Assume the reaction to go to completion.   1 L^.atm = 101.325 J           (7.39 x 10³J = 7.39 kJ, DU = -331.9 kJ)