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Chapter 6

Thermochemistry

Shaun Williams, PhD

The Nature of Energy

Energy

Ball on a Hill

Initial position - In the initial position, ball A has a higher potential energy than ball B. Ball A is held in place above the bottom of the hill. At the bottom of the hill is a stationary ball B.

Final position - After A has rolled down the hill, the potential energy lost by A has been converted to random motions of the components of the hill (frictional heating) and to the increase in the potential energy of B. Ball A is now stationary at the bottom of the hill and ball B has ascended the far side of the valley.

Energy

Chemical Energy

Thermodynamics

Internal Energy

The Sign of Internal Energy

A diagram that graphically shows the text given on this slide.

Work

  • \(\text{Work} = P × A × \Delta h = P\Delta V\)
    • \(P\) is pressure.
    • \(A\) is area.
    • \(\Delta h\) is the piston moving a distance.
    • \(\Delta V\) is the change in volume.
As a piston moved upwards in a cylinder, the change in height translates into a change in volume.

Work - cont.

Exercise 1

Which of the following performs more work?

  1. A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
  2. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

Exercise 1 - Answer

Which of the following performs more work?

  1. A gas expanding against a pressure of 2 atm from 1.0 L to 4.0 L.
  2. A gas expanding against a pressure of 3 atm from 1.0 L to 3.0 L.

They perform the same amount of work.

Enthalpy and Calorimetry

Change in Enthalpy

Exercise 2

Consider the combustion of propane: $$ \chem{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)} \;\;\; \Delta H=-2221\,\chem{kJ}$$ Assume that all of the heat comes from the combustion of propane. Calculate \(\Delta H\) in which 5.00 g of propane is burned in excess oxygen at constant pressure.

Exercise 2 - Answer

Consider the combustion of propane: $$ \chem{C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)} \;\;\; \Delta H=-2221\,\chem{kJ}$$ Assume that all of the heat comes from the combustion of propane. Calculate \(\Delta H\) in which 5.00 g of propane is burned in excess oxygen at constant pressure.

$$ -252\,\chem{kJ} $$

Calorimetry

A Coffee-Cup Calorimeter Made if Two Styrofoam Cups

A diagram of a coffee-cup calorimeter. The coffee cups have a stirred and a thermometer inserted into it.

Calorimetry - cont.

Hess's Law

\(\chem{N_2(g)+2O_2(g)\rightarrow 2NO_2(g)}\;\;\;\Delta H_1=68\,\chem{kJ}\)

This reaction also can be carried out in two distinct steps, with enthalpy changes designated by \(\Delta H_2\) and \(\Delta H_3\). $$\begin{align} \chem{N_2(g)+O_2(g)} &\rightarrow \chem{\cancel{2NO(g)}} \;\;\;\;\; & \Delta H_2=180\,\chem{kJ} \\ \chem{\cancel{2NO(g)}+O_2(g)} &\rightarrow \chem{2NO_2(g)} \;\;\;\;\; & \Delta H_3=-112\,\chem{kJ} \\ \hline \chem{N_2(g)+2O_2(g)} &\rightarrow \chem{2NO_2(g)} \;\;\;\;\; & \Delta H_2+\Delta H_3=68\,\chem{kJ} \end{align}$$ So we see that $$ \Delta H_1 = \Delta H_2+\Delta H_3 = 68\,\chem{kJ} $$

The Principle of Hess's Law

A diagram of the information presented on the previous slide.

Characteristics of Enthalpy Changes

Example

Consider the following data: $$\begin{align} \chem{NH_3(g)} &\rightarrow \chem{\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)} \;\;\; &\Delta H=46\,\chem{kJ} \\ \chem{2H_2(g)+O_2(g)} &\rightarrow \chem{2H_2O(g)} \;\;\; &\Delta H=-484\,\chem{kJ} \end{align}$$ Calculate \(\Delta H\) for the reaction $$ \chem{2N_2(g)+6H_2O(g)\rightarrow 3O_2(g)+4NH_3(g)} $$

Problem-Solving Strategy

Standard Enthalpies of Formation

Standard Enthalpies of Formation \(\left(\Delta H^\circ_f\right)\)

Conventional Definitions of Standard States

A Schematic Diagram of the Energy Changes for the Reaction $$\chem{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$

$$ \Delta H^\circ_\chem{reaction} = -\left(-75\,\chem{kJ}\right) + 0 + \left(-394\,\chem{kJ}\right) + \left(-572\,\chem{kJ}\right) = -891\,\chem{kJ} $$

Using the negative of the enthalpies of formation of the starting materials gives the enthalpy change to break up all the starting materials. Using the enthalpies of formation of the products give the enthalpy change to form the products. Summing up all of these gives the enthalpy change of the reaction.

Problem-Solving Strategy: Enthalpy Calculations

  1. When a reaction is reversed, the magnitude of \(\Delta H\) remains the same, but its sign changes.
  2. When the balanced equation for a reaction is multiplied by an integer, the value of \(\Delta H\) for that reaction must be multiplied by the same integer.
  3. The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products: $$ \Delta H^\circ_\chem{rxn} = \sum n_\chem{p} \Delta H^\circ_f\left(\chem{products}\right) - \sum n_\chem{r} \Delta H^\circ_f\left(\chem{reactants}\right) $$
  4. Elements in their standard states are not included in the \(\Delta H_\chem{reaction}\) calculations because \(\Delta H^\circ_f\) for an element in its standard state is zero.

Excercise 3

Calculate \(\Delta H^\circ\) for the following reaction: $$ \chem{2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)} $$ Given the following information:

\(\Delta H^\circ_f\,\left(\bfrac{\chem{kJ}}{\chem{mol}}\right)\)
\(\chem{Na(s)}\) 0
\(\chem{H_2O(l)}\) -286.
\(\chem{NaOH(aq)}\) -470.
\(\chem{H_2(g)}\) 0

Excercise 3 - Answer

Calculate \(\Delta H^\circ\) for the following reaction: $$ \chem{2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)} $$ Given the following information:

\(\Delta H^\circ_f\,\left(\bfrac{\chem{kJ}}{\chem{mol}}\right)\)
\(\chem{Na(s)}\) 0
\(\chem{H_2O(l)}\) -286.
\(\chem{NaOH(aq)}\) -470.
\(\chem{H_2(g)}\) 0
$$ \Delta H^\circ = -368.\,\chem{kJ} $$

Present Sources of Energy

Energy Sources Used in the United States

In 1850, 91% of our energy was from wood and 9% from coal. By 1900 21% was wood, 71% coal, petroleum 5% and hydro 3%. By 1950, 6% from wood, 36% from coal, 52% from petroleum, and 6% from hydro and nuclear. By 1975, 3% wood, 18% coal, 73% petroleum, and 6% hydro and nuclear. By 2000, 4% wood, 23% coal, 62% petroleum, and 11% hydro and nuclear.

The Earth's Atmosphere

Sumlight enters Earth's atmoshpere and strikes the surface heating the surface. The hot surface emits infrared radiation upward to cool itself. The infrared radiation should be able to leave the Earth's atmosphere. Carbon dioxide and water vapor absorb the infrared radiation and then reemit it. About 50% of that remission is directed downwards at the surface thereby effectively trapping that heat in the atmosphere.

New Energy Sources

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