
Chapter 6 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.

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.

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

• 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.

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

• 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.

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 $$\Delta H_1 = \Delta H_2+\Delta H_3 = 68\,\chem{kJ}$$

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:

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

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

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

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

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