Thermochemistry14.notebook November 24, 2015 · 2015. 11. 24. · Thermochemistry14.notebook 6 November 24, 2015 State of a System – refers to the values of all the pertinent macroscopic

Thermochemistry14.notebook

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Thermochemistry Energy the capacity to do work or produce heat

POTENTIAL ENERGY KINETIC ENERGY (energy of motion)

"stored" vibrationalbond energy rotational

translational

TEMPERATURE and HEAT

a measure of involves the transferaverage KE of energy between twoof particles objects due to a difference in temperature


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Chemical Energy – form of energy stored within the structural units of chemical substances

SYSTEM and SURROUNDINGS

the part of the universe that is the remainder of the universeof interest to us (all but the system)

open system – if it can exchange mass and energy with the surroundings

closed system – if it can exchange energy, but not mass, with the surroundings

isolated – if it cannot exchange mass or energy with the surroundings

Exothermic heat energy flows out of the system

Endothermic heat energy flows into the system


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PE Diagrams:

PE PE

exothermic endothermic∆H = ∆H = +

It is a feature of our universe that,

ENERGY LOST BY SYSTEM = ENERGY GAINED BY SURROUNDINGS (and vice versa)

The study of thermodynamics is concerned primarily with energy changes and the flow of energy from one substance to another. We will see that a more detailed analysis of these energy transfers allows us to understand why some changes are inevitable while others are impossible. Thermodynamics is organized around three fundamental, experimentally derived laws of nature. They are identified as the first, second, and third laws of thermodynamics.


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The First Law

The first law of thermodynamics deals with the exchanges of energy between a system ands its surroundings and is basically a statement of the law of conservation of energy. This law serves as the foundation for Hess’s law.

The first law can be stated mathematically as:

∆Euniv = ∆Esys + ∆Esurr = 0

While the energy of the universe never changes (∆Euniv = 0), the energy of the system and the energy of the surroundings can each change, and it will always be true that

∆Esys = ∆Esurr

The energy represented by Esys includes all forms of energy possessed by the system (e.g. thermal, potential, chemical, etc). It is often referred to as the internal energy of the system. We can never determine the value of the internal energy of a system because there is no natural reference point for energy. However, by monitoring the system and its interactions with the surroundings, we can measure changes in the internal energy.

ΔE = Efinal Einitial


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or, for a chemical reaction,

ΔE = Eproducts Ereactants

There are two ways in which the internal energy of a system may change:

1. For work (w) to be done

and/or

2. For heat (q) to flow

We can express this as

∆E = q + w

If work is done on a system, as in the compression of a gas, the system gains and stores energy. Conversely, if the system does work on the surroundings, as when a gas expands and pushes a piston, the system loses some energy by changing part of its potential energy to kinetic energy which is transferred to the surroundings.

q is (+) heat is absorbed by the systemq is () heat is released by the systemw is (+) work is done on the systemw is () work is done by the system


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State of a System – refers to the values of all the pertinent macroscopic properties of the system. For example, the state of an ideal gas can be can be completely specified by its volume, pressure, temperature and number of moles.

State Functions

These are properties that are determined by the state that the system is in (and not how it got there). Since the value of a sate function depends only upon the state that the system is in, the calculation of a change in that state function will depend only upon the initial and final states of the process (and not upon the “path”). Thus, we don’t need to know everything about the system throughout the process. All we need to know is the value of the state function of interest when it was in the initial state and the value of the state function of interest in its final state. Not all thermodynamicquantities are state functions. For example, heat (q) and work (w) depend on how we carry out a process.


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Work can be done by a gas through expansion or work can be done to a gas through compression. Consider a piston:

work = force x distance P = F/A

work = F x ∆h F = P x A

work = P x A x ∆h ∆V = A x ∆h

work = P x ∆V

work = P x ∆V

For a gas expanding, ∆V = + and therefore w = For a gas contracting, ∆V = and therefore w = +


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There are two kinds of work that chemical systems can do (or have done on them). One is electrical (which we will consider later in the year) and the other is work associated with the expansion or contraction of a system under the influence of an external pressure. Such pressurevolume work, or PV work is given by the equation

W = PΔV

where P is the external pressure on the system.

Now the first law of thermodynamics tells us that

∆E = q + w = q P∆V

Since we wish to monitor only heat, let’s see what we get when we solve this expression for q:

q = ∆E + P∆V

When a reaction takes place in a container whose volume cannot change (so that ΔV = 0), ΔE = q, which means the entire energy change must appear as heat that’s either absorbed or released. For this reason, ΔE is called the heat of reaction at constant volume (qv),

ΔE = qv (qv means heat at constant volume)


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Rarely do we carry out reactions in containers of fixed volume. Usually, reactions take place in containers open to the atmosphere where they are exposed to a constant pressure. To study heats of reactions under constant pressure conditions, the concept of enthalpy was invented. Enthalpy H is defined by the equation

H = E + PV

For a process at constant pressure,

∆H = ∆E + P∆V

Thus at constant pressure, both q and ∆H = ∆E + P∆V

And therefore ∆H = qp (qp means heat at constant pressure)

Enthalpy (H) is a state function.

Notice that ΔE and ΔH are not equal. They differ by the product PΔV; this can be seen by rearranging the above equation:

ΔH ΔE = PΔV

The ΔV values for reactions involving only solids and liquids are very tiny, so ΔE and ΔH for these reactions are nearly the same size. The only time ΔE and ΔH differ by a significant amount is when gases are formed or consumed in a reaction, and even then they do not differ by much.


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Enthalpy and Calorimetry

Calorimetry

Specific Heat Capacity (c) = the energy needed to raise one gram of a substance by one degree Celsius. The unit = J/goC

Molar Heat Capacity (c) = the energy needed to raise one mole of a substance by one degree Celsius. The unit = J/moloC

Constant pressure calorimeter = coffeecup calorimeter

∆H = q = mc∆T + ccalΔT

where ccal = coffee cup calorimeter constant (heat capacity of calorimeter); note the unit = J/oCm = mass of water in calorimeter; unit = (g)c = specific heat capacity; unit = (J/goC)∆T = change in temperature (Tf – Ti) and is always expressed as a positive value; unit = oC

The ccalΔT term in the above equation is often omitted in conventional textbook problems.


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Constant volume calorimeter = bomb calorimeter

V = 0 so, P∆V = 0 and thus w = 0

and ∆E = q + w = q

q = cbomb∆T

Cbomb = q/∆T = heat/oC


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Problem 1: A balloon contains 313g He at a pressure of 1.00 atm. The volume of the balloon is 1910 L. The temperature is decreased by 15oC as the volume decreases to 1814 L, the pressure remaining constant. Calculate q, w, and ΔE for thehelium in the balloon. The molar heat capacity of helium gas is 20.8 J/moloC. Note: 1 Latm = 101.3 J.


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Problem 2:Exactly 3.358 kJ of heat is added to a calorimeter that contains 50.00 g of water. The temperature of the water and calorimeter, originally at 22.34oC, increases to 36.74oC. Calculate the heat capacity of the calorimeter in J/oC


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Problem 3: A 50.00 mL sample of 0.500 HCl solution at 23.35oC is mixed with 50.00 mL of 0.500 M NaOH solution, also at 23.35oC, in the coffee cup calorimeter used in the previous problem. After the reaction occurs, the temperature of the resulting mixture is measured to be 26.82oC. The density of the final solution is 1.02 g/mL. Calculate ΔH for the reaction. Assume the specific heat of the solution is the same as that of water.


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Problem 4: A 1.000. g sample of ethanol, C2H5OH, was burned in a bomb calorimeter whose heat capacity had been determined to be 2.71 kJ/oC. The temperature rose from 24.284oC to 34.727oC. Determine ΔE for the reaction in joules per gram of ethanol, and also in kilojoules per mole of ethanol. Write a balanced equation for this reaction.


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Hess's Law in going from a particular set of reactants to a particular set of products, the enthalpy change is the same whether the reaction takes place in one step or a series of steps (enthalpy is a state function).

Keep in Mind the Following Key Concepts When Doing Enthalpy Calculations:

1. When a reaction is reversed, the magnitude of ∆H remains the same, but its sign changes

2. When the balanced equation for a reaction is multiplied by an integer, the value of the H for that reaction must be multiplied by the same integer (i.e. ∆ H is an extensive property)

3. The change in ethalpy for a given reaction can be calculated from the standard enthalpies of formation of the reactants and products:

∆Horeaction = Σ∆Hf(products) Σ∆Hf(reactants)

standard states:

for a gas, P = 1 atmfor a solution, 1.0 Mfor condensed state, pure liquid or solidfor an element, standard state is the form in which the element exists under standard conditions of 1 atm and 25oC


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Problem 5: Given the following data

N2(g) + O2(g) à 2NO(g) ΔH = +180.7 kJ

2NO(g) + O2(g) à 2NO2(g) ΔH = 113.1 kJ

2N2O(g) à 2N2(g) + O2(g) ΔH = 163.2 kJ

Use Hess’ law to calculate ΔH for the reaction

N2O(g) + NO2(g) à 3NO(g)


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Other Important Considerations:

Fossil Fuels

Petroleum viscous liquid mixture composed mostly of hydrocarbons

Fractional Distillation separating components in a mixture like petroleum on the basis of difference in boiling points.

Natural Gas mostly methane (CH4) found in petroleum deposits

Coal remains of plants buried and subjected to high pressure and heat over long periods of time

fuels that consist of carbonbased molecules derived from the decomposition of once living organisms; coal, petroleum, and natural gas


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Energy Alternatives


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Thermochemistry14.notebook November 24, 2015 · 2015. 11. 24. · Thermochemistry14.notebook 6 November 24, 2015 State of a System – refers to the values of all the pertinent macroscopic

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