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Energy and Chemical Reactions - ThermochemistryAlkali Metals in Water: http://www.youtube.com/watch?v=uixxJtJPVXk&feature=related

Key Concepts-Our Goals are to:1. develop an understanding of Potential and Kinetic Energy and the ability to describe the relationship

between the two in Thermochemistry.

2. be able to define the terms System and Surroundings and to identify a System and its Surroundings for physical and chemical changes.

3. develop an understanding of the concept of Heat Transfer and the ability to describe the process.

4. develop an understanding of the concept of Pressure-Volume Work and the ability to recognize when a system can do work or when a system has work done on it.

5. develop the ability to apply and use the First Law of Thermodynamics.

6. develop an understanding of internal energy change for a system, ∆Esys; be able to relate this to heat flow and work (Learn the sign conventions!).

7. develop an understanding of the concept of a State Function; know what variables in chemistry are State Functions.

8. develop an understanding of ENTHALPY Change, ∆H; know that this is heat transfer at constant pressure, a State Function.

9. develop an understanding of Enthalpies of Reaction, ∆Hrxn; be able to apply Hess’s Law to determine ∆Hrxn and to calculate ∆Hrxn from measured heat flow.

10. develop an understanding of Enthalpies of Formation, ∆H°f, and the ability to write reactions corresponding to standard enthalpies of formation.

11. develop the ability to do stoichiometric calculations that include heats of reaction.

12. learn the basics of Calorimetry; be able to use calorimetry data to calculate ∆Hrxn, specific heat capacity, temperature change, amount of heat flow, etc..(Covered in LAB only.)

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Energy-Kinetic vs. Potential EnergyEnergy - The capacity to do work. Common units are Joules and Calories

Joule (J): SI unit of energy (A very small amount of energy so we commonly use the kilojoule, kJ.)1 J = 1 Newton•m = 1 kg•m2/s2 calorie (cal): 1 cal = 4.184 J (The heat required to raise the temperature of 1 g of water from 14.5°C to 1 5.5°C.)Nutritional calorie (Cal) = 1 kcal = 1000 cal

Kinetic Energy: Energy due to motion of matter.In chemistry we are often concerned with the thermal (heat) energy of atoms/molecules/ions. Heat is a manifestation of random kinetic energy. Heat is directly proportional to temperature.

Heat and Temperature are not the same!Heat is a measure of energy content.Heat is an extensive property.

Temperature is a measure of AVERAGE kinetic energy.Temperature is an intensive property.

Heat is always transferred from an object with a higher temperature(higher average kinetic energy of the particles) to one of lowertemperature (lower average kinetic energy of the particles) untilTHERMAL EQUILIBRIUM is reached.

Potential Energy: Stored energy in a system. Weakly temperature dependent.Chemical Potential Energy-The kind of Potential Energy of interest in Thermochemistry. Chemical potential energy results from attractive and repulsive forces between nuclei and electrons in a system. Includes the bond energies between atoms bonded together within a molecule (intramolecular), forces between ions, and forces between different molecules (intermolecular) in liquids and solids.

Amount of Heat Lost = Amount of Heat Gained

Evisdom

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Kinetic EnergyKinetic Energy (KE)To calculate the kinetic energy of a moving object in joules, what units must you use for the object’s mass, m, and velocity, u?

Text Question 5.16:a) A baseball weighs 5.13 oz. What us the kinetic energy in joules of this baseball when it

is thrown by a major-league pitcher at 95.0 mph?

b) By what factor will the kinetic energy change if the speed of the baseball is decreased to 55.0 mph?

c) What happens to the kinetic energy when the baseball is caught by the catcher?

KE = 12mu2

When we study chemical systems we actually study ENERGY TRANSFER NOT ENERGY CONTENT.

Energy is TRANSFERRED between a SYSTEM and SURROUNDINGS as heat (q) and/or work (w).System: Process under study (chemical reaction or physical change).

Surroundings: Everything else. May be isolated as in a calorimetry experiment.

Closed System: A system that can exchange energy, but not matter with its surroundings.

Universe = System + Surroundings

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Thermochemistry Basic Terms

SystemUniverse

heat qwork w

The Internal Energy (E) of the SYSTEM: The sum of all the kinetic and potential energies of the matter in the system. CANNOT BE absolutely determined.

However, we can easily measure changes in internal energy (∆Esys) for a system:

∆Esys = Efinal - Einitial = q + w

Sign conventions:

1. q (+) HEAT flows into the system from the surroundings. ENDOTHERMIC PROCESS

2. q (-) HEAT flows out of the system to the surroundings. EXOTHERMIC PROCESS

3. w(+) WORK is done ON the system by the surroundings.

4. w(-) WORK is done BY the system on the surroundings.

∆ Esys (+) SYSTEM internal energy INCREASES.

∆ Esys (−) SYSTEM internal energy DECREASES.

First Law of Thermodynamics:

Energy can be transferred from one form to another, but cannot be created or destroyed in a physical or chemical change.

Surroundings

∆ Euniv = ∆ Esys +∆ Esurr = 0

∆ Esys = −∆ Esurr

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Definitions and Sign Conventions for Work (w)Work (w): Energy that moves an object against a force is called work. Work is quantified as force x distance:

Only pressure-volume work will be considered here. For a gaseous system expanding or contracting in volume (distance) of ∆V = Vfinal – Vinital, against a constant external pressure (force), P, the work is defined as:

SI units are pascals for pressure and m3 for volume. Chemists like to use atmospheres for pressure and liters for volume. To convert to Joules we use 1 L•atm = 101.325 J.

wsys > 0, ∆Esys increasesWork is done on the system. This is the case when the volume of a gas system decreases, ∆V is negative, so w = –P∆V is positive.

wsys < 0, ∆Esys decreases Work is done by the system. This is the case when the volume of a gas system increases, ∆V is positive, so w = –P∆V is negative.

w = F • d

w = F • d = −P(∆V )

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Example Work for a Physical Change: Solid —> GasReaction systems that result in a change in moles of gas experience a volume change. An example where work is done by the system is sublimation of CO2: CO2(s) —> CO2(g)

Another example is the denotation of TNT.Here 15 moles of gas are created for every 2 moles (450 g) of TNT. Assuming a temperature of 525°C, the hot expanding gasses can do ≈100 kJ of work. Can you think of a simple example where wsys > 0?

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State Functions in ChemistryState functions are functions whose CHANGE in value depends only upon the the initial and final states of the system, not on the pathway from initial to final.

∆(State function) = Final State - Initial State

By convention, all state functions are represented by UPPERCASE letters in chemistry.

∆T (temperature), ∆P (pressure), ∆V (volume), ∆E (internal energy) are all state functions.

Variables that are not state functions: w (work), d (distance), t (time),q (heat), etc.

Note: the sum q + w = ∆E is a state function.

The relative amounts of heat and work for a given ∆E depends upon the pathway over which the energy change occurs. Consider the discharge of the potential energy stored in a battery. We can vary the proportions of heat to work by changing the circuitry attached to the battery. The same amount of energy is used in both cases, but the pathway of energy use is different.

A Special State Function: ENTHALPY, HAt constant external pressure, ∆P = 0, the heat transferred between the system and surroundings is called enthalpy (H). The change in enthalpy is defined as:

∆Hsys is the Enthalpy of Reaction(Heat of reaction at constant P, a STATE FUNCTION.)

∆Hsys = ∆Ep + P∆V = ?

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Heat Transfer at Constant Pressure, ∆HNote how changes in internal energy and changes in enthalpy (heat content) at constant pressure are related, differing only in the quantity of energy transferred to or from the system as work:

∆E = ∆H – P∆V = ∆H + w (Remember w = – P∆V)

For many processes ∆V is very small, or negligible, making ∆E ≈ ∆H.What are some examples of processes with negligible ∆V?(Hint: What processes DO NOT involve a change in the number of moles of gas, thus ∆V ≈ 0?)

In processes where ∆V is large, ∆E and ∆H will be different.What are some examples of processes with significant ∆V?(Hint: What processes DO involve a change in the number of moles of gas, thus ∆V ≠ 0?)

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Example Chemical SystemConsider the redox reaction of zinc metal with hydrochloric acid: Zn(s) + 2 HCl(aq) –> ZnCl2(aq) + H2(g)Consider the case where this reaction is done at a constant external pressure of 1.00 atm and T = 25°C.

For the complete reaction of 1 mole of zinc:

1. The amount of heat absorbed by the surroundings is 154 kJ.

2. The volume of H2 gas generated is 24.5 liters.

We symbolize and quantify this as follows:

1. qp = ∆Hsys = –154 kJ (Exothermic)

2. ∆Hsurr = – ∆Hsys = +154 kJ

3. w = -P(∆V) = (-1.00 atm)(+24.5 L) = -24.5 L*atm = -2.48 kJ

4. ∆Esys = ∆Hsys + w = (–154 + -2.48) kJ = -156 kJ

5. ∆Esurr = – ∆Esys = +156 kJ

Observations:

The chemical reaction is EXOTHERMIC. The surroundings will gain heat and should show an increase in T.

The SYSTEM loses 154 kJ of POTENTIAL energy (not thermal energy) to the surroundings.

The SYSTEM does 2.48 kJ of work on the surroundings.

In this chemical system, |∆Hsys| >> |w| making ∆Esys ≈ ∆Hsys.

SY

ST

EM

Ent

halp

y, ∆

Hsy

s

–154 kJ

Zn(s) + 2 HCl(aq)

ZnCl2(aq) + H2(g)

Initial State of SYSTEM

Final State of SYSTEM

Enthalpy Diagram

Work

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∆H of Reactions, ∆Hsys

CH4(g) + 2O2(g) —> CO2(g) + 2H2O(l) ∆Hrxn = –890 kJ

For the above reaction the value of ∆Hsys represents the difference in chemical potential energy between the products (final state) and the reactants (initial state)

∆Hsys = H(products) - H(reactants)

So if ∆Hsys is a difference in potential energy then why do we refer to it as a heat? Because we measure the system’s difference in potential energy (∆Hsys) by measuring the HEAT change of the surroundings (qsurr) when the chemical reaction takes place.

∆(potential energysys) = ∆(heatsurr)

Thus, ∆Hsys = -qsurr. In the surroundings we measure a heat change by experimentally observing a change in temperature. This is the basis for calorimetry (covered in lab).

Energy of ReactionsDaley/Larson 11

Review of Energy Changes

Universe

System Surroundings

heat qwork w

Potential Energy Changes

(Temperature constant)

Kinetic Energy Changes

(Temperature changes)

Make sure understand the changes in the system versus the surroundings when we study chemical reactions and physical state changes. The “system” undergoes potential energy changes when the reactants are transformed into products or when a physical state change occurs. The temperature of the system is taken as the same before and after reaction, no thermal energy change in the system. Heat flow to or from the surroundings maintains the systems temperature.

The surroundings are a reservoir of thermal energy where heat is absorbed or released. Any potential energy change in the system is reflected by a corresponding thermal energy change in the surroundings. This is what we can “see” (measure) in the laboratory. How do we know when a potential energy change takes place in a system? We observe the effects of the system’s potential energy change on the surroundings because the surroundings experience a change in temperature. This is the basis for calorimetry.

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Heat of Reactions, ∆Hrxn

For chemical reactions we measure or calculate ∆Hsys for a balanced chemical reaction.

CH4(g) + 2O2(g) —> CO2(g) + 2 H2O(l) ∆Hrxn = –890 kJ = ∆Hsys

1. ∆Hrxn is specific to the reaction given with the coefficients given.

2. The physical states (s, l, g, aq) are required and important!

3. The coefficients in the balanced equation determine the extent of ∆Hrxn.∆Hrxn therefore depends on the amount of reaction, and is an EXTENSIVE property. If you multiply the coefficients of a balanced equation by a constant, ∆Hrxn is multiplied by the same constant.

4. If a chemical reaction is reversed, then the sign of ∆Hrxn changes but the magnitude remains the same. CO2(g) + 2H2O(l) —> CH4(g) + 2 O2(g) ∆Hrxn = +890 kJ

5. For an exothermic system we think of heat as a product of the reaction: reactants —> products + heat (Thermodynamically favorable, but can still be non-spontaneous for certain systems under certain conditions. This will be explored in more detail in chemistry 1B.)

6. For an endothermic system we think of heat as a reactant of the reaction: heat + reactants —> products(Thermodynamically unfavorable, but can still occur spontaneously for certain systems under the right conditions. This will be explored in more detail in chemistry 1B.)

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Stoichiometry and ∆Hrxn

For the reaction:N2(g) + 3 H2(g) —> 2 NH3(g) ∆Hrxn = –91.80 kJ (so heat is a product, exothermic)

We can interpret the magnitude of ∆Hrxn in many ways:

1. 91.80 kJ of heat is produced when 1 mol of N2 reacts or

2. 91.80 kJ of heat is produced when 3 mol of H2 reacts or

3. 91.80 kJ of heat is produced when 2 mol of NH3 is formed

1. What is ∆H/mol N2(g), ∆H/mol H2(g), ∆H/mol NH3(g)?

2. How much heat is produced if 23.6 g of hydrogen reacts?

3. How many grams of ammonia are produced when 750 kJ of heat is given off?

4. For the system, is the work term + or –?

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Finding ∆Hrxn from a Measured Heat TransferGiven a measured heat transfer for a reaction (qsurr) we can determine the ∆Hrxn.

Experiment shows 1.14 kJ of heat is needed to decompose 1.50 g of NaHCO3(s).

2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g) ∆Hrxn = ?

From the data, find ∆Hrxn.

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Hess’s Law of Heat SummationHess’s Law states that if a reaction is the sum of two or more other reactions, then the net ∆Hrxn is the sum of the individual ∆Hrxn values.

A —> B + C ∆H1

B —> D ∆H2 A —> C + D ∆H1 + ∆H2 = ∆Hrxn

In other words, the overall change in enthalpy (∆H) for a process is independent of the path by which the change occurs. Why is this true?

Consider the following reactions:1. 2 NaHCO3(s) —> Na2CO3(s) + H2O(g) + CO2(g); ∆H1 = 128 kJ

2. Na2CO3(s) —> Na2O(s) + CO2(g); ∆H2 = 138 kJ

3. 2 NaHCO3(s) —> Na2O(s) + H2O(g) + 2 CO2(g); ∆H3 = 266 kJ = ∆H1 + ∆H2

Reaction (3) is the sum of reactions (1) and (2). If reaction 1 is carried out, followed by (2), the overall enthalpy change will be equal to the enthalpy change for reaction (3) alone.

Bottom-line: For any reaction that can be written as the sum of simpler reactions, ∆Hrxn is the sum of the heats from the individual reactions.

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Using Hess’ LawUse Hess’ Law to find the unknown ∆Hrxn.

1. 2 Al(s) + 3/2 O2(g) —> Al2O3(s) ∆Hrxn = –1669.8 kJ

2. Al(s) + 3/2 Cl2(g) —> AlCl3(s) ∆Hrxn = –705.6 kJ

3. H2(g) + 1/2 O2(g) —> H2O(g) ∆Hrxn = –241.82 kJ

4. 1/2 N2(g) + 2 H2(g) + 1/2 Cl2(g) + 2 O2(g)—> NH4ClO4(s) ∆Hrxn = –295.3 kJ

5. 6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s) ∆Hrxn = ?

1. Work with only the reactants and products given in the target reaction. (All other species should cancel when summed.)

2. Reverse equations where reactants or products are on the incorrect side. Change the sign of ∆H.

3. Get the correct stoichiometric amounts (correct coefficients) of reactants and products on each side by multiplying each equation by the appropriate factor. Multiply ∆H by the same factor. (Sometimes fractions are needed.)

4. Add the reactions together, canceling species with the same formula and phase that are on opposite sides of the reaction.

5. Check that you have obtained the target reaction.

6. Sum the ∆H’s.

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Standard Enthalpies of Formation, ∆H°f

Standard Enthalpy of Formation (∆Hf°), or Heat of Formation.1. The enthalpy change for the formation of one mole of a compound (the product) directly from its component

elements (the reactants) in their standard states.

2. The standard state of an element is its most stable form in the physical state it exists at a pressure of 1bar (about 1atm) and at a specified temperature.

3. Data for the heat of formation of almost all compounds has been measured and/or calculated under Standard Thermochemical Conditions.

3.1. Standard State Thermochemical Conditions:1 bar pressure ≈ 1 atm25° C = 298.15 KAqueous Solutions: 1MPure elements: most stable form at 25° C and 1 bar.Example: C(s) is more stable as C(s,graphite) not C(s,diamond).

C(s,graphite) is used for a formation reaction involving carbon.

Examples:

H2(g) + 1/2 O2(g) → H2O(l) ∆Hf˚ = –285.8 kJ/mol

H2(g) + 1/2 O2(g) → H2O(g) ∆Hf˚ = –241.8 kJ/mol

2 C(graphite) + 3 H2(g) + 1/2 O2(g) → CH3CH2OH(l) ∆Hf˚ = –277.7 kJ/mol

Write a standard enthalpy of formation reaction for Na2CO3(s).

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Notes on Standard Enthalpies of Formation, ∆H°f

1. The ∆H°f for all elements in their standard states are zero. Elements are not formed in nature, they already exist.

2. Most ∆H°f values are negative indicating most compounds are more “stable” than their elements.

3. ∆H°f can be used to compare stabilities of compounds. Generally, the more negative ∆H°f the more “stable”.

4. For aqueous compounds, the values refer to 1 M solutions being formed. This includes any energy changes that occur during hydration of ions.

5. For some ionic compounds dissolved in water, you may only be able to find ∆Hf˚ values for the separate aqueous ions. For example, for FeCl3(aq) you may not find the ∆Hf˚ value for FeCl3(aq), but may find the ions separately:Fe3+(aq) and Cl–(aq). (In other words, you may need to use net ionic equations in some cases when you use ∆Hf˚ values.)

Why is H2O(s) not listed?

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Hess’ Law and ∆H°rxn

We can use tabulated values of ∆H°f to calculate the standard heat of reaction, ∆H°rxn, for any possible reaction.Use Appendix C in the textbook to find ∆H°f values.

We apply Hess’ Law:

Example: Find ∆H°rxn for the following:

6 NH4ClO4(s) + 10 Al(s) —> 3 N2(g) + 12 H2O(g) + 4 Al2O3(s) + 2 AlCl3(s)∆H°f (kJ/mol) –295.3 0 0 –241.82 –1669.8 –705.6 Look-up values using reference source.

∆H°f (kJ) –1771.8 0 0 –2901.84 –6679.2 –1411.2 Mult. by the molar stoichiometric coeff.

∑∆H°f (kJ) –1771.8 (reactants) –10992.2 (products) Sum products and reactants separately.

∆H°rxn (kJ) = –10992.2 – (–1771.8) = -9220.4 kJ Subtract reactant sum from product sum.

∆Hrxn

o = n∆H fo∑ ( products ) − n∆H f

o∑ ( reactants )

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Using ∆H°f to Calculate ∆H°rxn

Use Appendix C and Hess’ Law to find ∆H°rxn for each of the following:

1. H2O(l) —> H2O(g)

2. Fe2O3(s) + 2 Al(s) —> 2 Fe(s) + Al2O3(s)

3. NH4Cl(aq) + AgNO3(aq) —> AgCl(s) + NH4NO3(aq)

Thermite Welding

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Text ProblemsProblem 5.9: Imagine a container placed in a tub of water, as depicted in the following diagram:

(a) If the contents of the container are the system and heat is able to flow through the container walls, what qualitative changes will occur in the temperatures of the system and in its surroundings?

What is the sign of q associated with each change?

From the system’s perspective, is the process exothermic or endothermic?

(b) If neither the volume nor the pressure of the system changes during the process, how is the change in internal energy related to the change in enthalpy?

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Text ProblemsProblem 5.29: A gas is confined to a cylinder fitted with a piston and an electrical heater, as shown in the accompanying illustration:

Suppose that current is supplied to the heater so that 100 J of energy is added. Consider two different situations. In case (1) the piston is allowed to move as the energy is added. In case (2) the piston is fixed so that it cannot move.

(a) In which case does the gas have the higher temperature after addition of the electrical energy? Explain.

(b) What can you say about the values of q and w in each of these cases?

(c) What can you say about the relative values of ∆E for the system (the gas in the cylinder) in the two cases?

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Text ProblemsProblem 5.104a: Calculate the standard enthalpy of formation of gaseous diborane (B2H6) using the following thermochemical equations:

4B(s) + 3O2(g) —> 2B2O3(s) ∆H° = –2509.1 kJ 2H2(g) + O2(g) —> 2H2O(l) ∆H° = –571.7 kJ B2H6(g) + 3O2(g) —> B2O3(s) + 3H2O(l) ∆H° = –2147.5 kJ

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Text ProblemsProblem 5.113: Consider the following unbalanced oxidation-reduction reactions in aqueous solution:

(a) Balance each of the reactions.(b) By using data in Appendix C, calculate ∆H° for each of the reactions.(c) Based on the values you obtain for which of the reactions would you expect

to be thermodynamically favorable? Which would you expect to be unfavorable?

(d) Use the activity series to predict which of these reactions should occur. Are these results in accord with your conclusion in part (c) of this problem?

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Text ProblemsProblem 5.86: The heat of combustion of ethanol, C2H5OH(l), is −1367 kJ/mol. A batch of Sauvignon Blanc wine contains 10.6% ethanol by mass. Assuming the density of the wine to be 1.0 g/mL, what caloric content does the alcohol (ethanol) in a 6-oz glass of wine (177 mL) have?

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Text ProblemsProblem 5.102: Meals-Ready-to-Eat (MREs) are military meals that can by heated on a flameless heater. The heat is produced by the following reaction: Mg(s) + 2H2O(l) —> Mg(OH)2(s) + H2(g)Calculate the number of grams of Mg needed for this reaction to release enough energy to increase the temperature of 75 g of water from 21°C to 79°C.