Chapter 5

Thermochemistry

Shaun Williams, PhD

The Nature of Energy

Energy

• Capacity to do work or to produce heat.
• Law of conservation of energy – energy can be converted from one form to another but can be neither created nor destroyed.
  • The total energy content of the universe is constant.
• Potential energy – energy due to position or composition.
• Kinetic energy – energy due to motion of the object and depends on the mass of the object and its velocity.

Ball on a Hill

Initial position - In the initial position, ball A has a higher potential energy than 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.

Energy

• Heat involves the transfer of energy between two objects due to a temperature difference.
• Work – force acting over a distance.
• Energy is a state function; work and heat are not
  • State Function – property that does not depend in any way on the system’s past or future (only depends on present state).

Chemical Energy

• System – part of the universe on which we wish to focus attention.
• Surroundings – include everything else in the universe.
• Endothermic Reaction:
  • Heat flow is into a system.
  • Absorb energy from the surroundings.
• Exothermic Reaction:
  • Energy flows out of the system.
  • Energy gained by the surroundings must be equal to the energy lost by the system.

Thermodynamics

• The study of energy and its interconversions is called thermodynamics.
• Law of conservation of energy is often called the first law of thermodynamics.

Internal Energy

• Internal energy \(E\) of a system is the sum of the kinetic and potential energies of all the "particles" in the system.
• To change the internal energy of a system: $$ \Delta E = q + w $$
  • \(q\) represents heat
  • \(w\) represents work
• Thermodynamic quantities consist of two parts:
  • Number gives the magnitude of the change.
  • Sign indicates the direction of the flow.

The Sign of Internal Energy

• Sign reflects the system’s point of view.
• Endothermic Process:
  • \(q\) is positive
• Exothermic Process:
  • \(q\) is negative

• System does work on surroundings:
  • \(w\) is negative
• Surroundings do work on the system:
  • \(w\) is positive

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.

Work - cont.

• For an expanding gas, \(\Delta V\) is a positive quantity because the volume is increasing. Thus \(\Delta V\) and \(w\) must have opposite signs: $$ w=-P\Delta V $$
• To convert between \(\chem{L\cdot atm}\) and \(\chem{Joules}\), use \(1\,\chem{L\cdot atm} = 101.3\,\chem{J}\).

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.

Enthalpy and Calorimetry

Change in Enthalpy

• State function
• \(\Delta H = q\) at constant pressure
• \(\Delta H = H_\chem{products} – H_\chem{reactants}\)

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.

Calorimetry

• Science of measuring heat
• Specific heat capacity:
  • The energy required to raise the temperature of one gram of a substance by one degree Celsius.
• Molar heat capacity:
  • The energy required to raise the temperature of one mole of substance by one degree Celsius.
• If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction taking place is exothermic.
• An endothermic reaction cools the solution.

A Coffee-Cup Calorimeter Made if Two Styrofoam Cups

Calorimetry - cont.

• \(\text{Energy released (heat)} = s\times m \times \Delta T\)
  • \(s\) - specific heat capacity (\(\bfrac{\chem{J}}{\chem{{}^\circ C\cdot g}}\))
  • \(m\) - mass of solution (\(\chem{g}\))
  • \(\Delta T\) - change in temperature (\({}^\circ \chem{C}\))

Hess's Law

• In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

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

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

Characteristics of Enthalpy Changes

• If a reaction is reversed, the sign of \(\Delta H\) is also reversed.
• The magnitude of \(\Delta H\) is directly proportional to the quantities of reactants and products in a reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of \(\Delta H\) is multiplied by the same integer.

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

• Work backward from the required reaction, using the reactants and products to decide how to manipulate the other given reactions at your disposal.
• Reverse any reactions as needed to give the required reactants and products.
• Multiply reactions to give the correct numbers of reactants and products.

Standard Enthalpies of Formation

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

• Change in enthalpy that accompanies the formation of one mole of a compound from its elements with all substances in their standard states.

Conventional Definitions of Standard States

• For a Compound
  • For a gas, pressure is exactly 1 atm.
  • For a solution, concentration is exactly 1 M.
  • Pure substance (liquid or solid)
• For an Element - The form [\(\chem{N_2(g)}\), \(\chem{K(s)}\)] in which it exists at 1 atm and 25°C.

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

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

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:

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:

Present Sources of Energy

• Fossil Fuels
  • Petroleum, Natural Gas, and Coal
• Wood
• Hydro
• Nuclear

Energy Sources Used in the United States

The Earth's Atmosphere

• Transparent to visible light from the sun.
• Visible light strikes the Earth, and part of it is changed to infrared radiation.
• Infrared radiation from Earth’s surface is strongly absorbed by \(\chem{CO_2}\), \(\chem{H_2O}\), and other molecules present in smaller amounts in atmosphere.
• Atmosphere traps some of the energy and keeps the Earth warmer than it would otherwise be.

New Energy Sources

• Coal Conversion
• Hydrogen as a Fuel
• Other Energy Alternatives
  • Oil shale
  • Ethanol
  • Methanol
  • Seed oil