22.5: Gibb's Free Energy
Student Behavioral Objectives
The student will:
- define Gibbs Free Energy.
- calculate Gibbs Free Energy given the enthalpy and entropy.
- use Gibbs Free Energy to predict spontaneity.
Timing, Standards, Activities
Lesson | Number of 60 min periods | CA Standards |
---|---|---|
Gibb's Free Energy | 1.5 | 7f |
Activities for Lesson 5
Laboratory Activities
1. None
Demonstrations
1. None
Worksheets
1. Enthalpy, Entropy, and Free Energy Worksheet
Extra Readings
1. None
Answers for Gibb's Free Energy (L5) Review Questions
- Sample answers to these questions are available upon request. Please send an email to teachers-requests@ck12.org to request sample answers.
Multimedia Resources for Chapter 22
From the Annenberg videos . . . Exothermic and Endothermic reactions.
This video provides a blackboard lecture showing the concepts involved with an example of using Hess's Law.
The learner.org website allows users to view streaming videos of the Annenberg series of chemistry videos. You are required to register before you can watch the videos but there is no charge. The website has one video that relates to this lesson called “The Driving Forces.”
This video shows an example of how to plug values into Gibbs Free Energy equation.
A blackboard example of using Gibbs free energy equation to determine whether a reaction is spontaneous.
Laboratory Activities for Chapter 22
Teacher's Resource Page for Heat of Reaction - Hess's Law
Lab Notes
Solid sodium hydroxide is highly hygroscopic . . . it absorbs water from the air. The solution produced by this absorption is extremely corrosive . . . it will damage clothes almost immediately even though the wearer may not notice for several hours. If the students leave the lid off the reagent bottle after removing \begin{align*}NaOH\end{align*} pellets, the \begin{align*}NaOH\end{align*} will absorb enough water to ruin the bottle of reagent. The \begin{align*}NaOH\end{align*} pellets may appear to only be slightly damp but later they will become one solid rock and unusable. Also, any pellets spilled on the table or floor will become a serious hazard due to slippery floors and corrosive puddles for hands and sleeves to be dipped into. Spills must be cleaned up immediately and jar lids must be closed immediately. Direct and constant supervision of this step is desirable.
The \begin{align*}200. \ mL\end{align*} of water used by the students in step 1 may be tap water. But when tap water comes out of the faucet, it is considerably colder than room temperature. Using very cold water will encourage heat absorption from the air during the reaction and will affect the \begin{align*}\Delta t\end{align*}. For this reason, the water should be taken from the tap at least one day ahead of the lab and left in the room to adjust to room temperature. It can be stored in used (but clean) milk containers. For 15 pairs of students, you will need about \begin{align*}4 \ liters\end{align*} of water.
Solution Preparation
For 15 pairs of students, you will need about \begin{align*}2 \ liters\end{align*} of \begin{align*}0.50 \ M\end{align*} \begin{align*}NaOH\end{align*} solution. This can be prepared by dissolving \begin{align*}20.00 \ grams\end{align*} of solid \begin{align*}NaOH\end{align*} per liter of solution. It is advised to fill the volumetric flask half full of distilled water, add the solid \begin{align*}NaOH\end{align*}, swirl to dissolve the solid, allow some time for the solution to cool (the dissolving of \begin{align*}NaOH\end{align*} is highly exothermic), then fill the volumetric flask to the \begin{align*}1.00 \ liter\end{align*} mark.
For 15 pairs of students, you will need about \begin{align*}2 \ liters\end{align*} of \begin{align*}0.50 \ M\end{align*} \begin{align*}HCl\end{align*} solution. This can be prepared by dissolving \begin{align*}42 \ mL\end{align*} of concentrated \begin{align*}HCl\end{align*} \begin{align*}(12 \ M)\end{align*} per liter of solution. Always add acid to water not water to acid. Again, fill the volumetric flask half full of distilled water, add the concentrated \begin{align*}HCl\end{align*}, swirl to mix, and fill to the \begin{align*}1.00 \ liter\end{align*} mark.
For 15 pairs of students, you will need about \begin{align*}4 \ liters\end{align*} of \begin{align*}0.25 \ M\end{align*} \begin{align*}HCl\end{align*}. When you prepare the \begin{align*}0.50 \ M \ HCl\end{align*} (see above) make 2 extra liters and then dilute these by half to make \begin{align*}0.25 \ M\end{align*} \begin{align*}HCl\end{align*}.
Disposal
The solutions produced in this lab can be disposed of down the sink with excess water.
Solutions to Pre-Lab Questions
- \begin{align*}q = mC \Delta t = (500. \ g)(4.18 \ J/g \cdot ^\circ C)(3.5^\circ C) = 7315 \ J = 7.3 \ kJ\end{align*}
- Diluting a solution does not change the moles of solute contained in it. If the moles of \begin{align*}NaCl\end{align*} in the original \begin{align*}100. \ mL\end{align*} is \begin{align*}0.0100 \ moles\end{align*}, then the moles of \begin{align*}NaCl\end{align*} in the \begin{align*}200. \ mL\end{align*} of diluted solution will also be \begin{align*}0.0100 \ moles\end{align*}.
Solutions to Post-Lab Questions
1. (a) Since the grams of \begin{align*}NaOH\end{align*} is doubled, the heat given off will also double.
(b) Since the grams of \begin{align*}NaOH\end{align*} is doubled, the moles would also double.
(c) If the heat is doubled and the moles are doubled, heat per mole would remain exactly the same.
Heat of Reaction - Hess's Law
Background:
Hess's Law; The enthalpy change for any reaction depends on the products and the reactants and is independent of the pathway or the number of steps between the reactant and the product.
In this experiment, you will measure and compare the quantity of heat involved in three reactions. These heats of reaction will be measured using a styrofoam calorimeter. The three reactions are shown below.
Reaction 1: The dissolving of solid sodium hydroxide in water.
\begin{align*}NaOH_{(s)} \rightarrow Na^+_{(aq)}+ OH^-_{(aq)} \qquad \Delta H_1 =\end{align*}
Reaction 2: The reaction of solid sodium hydroxide with dilute hydrochloric acid.
\begin{align*}NaOH_{(s)}+ H^+_{(aq)}+ Cl^-_{(aq)} \rightarrow Na^+_{(aq)}+ Cl^-_{(aq)}+ H_2O_{(L)} \qquad \Delta H_2 =\end{align*}
Reaction 3: The reaction of sodium hydroxide solution with dilute hydrochloric acid.
\begin{align*}Na^+_{(aq)}+ OH^-_{(aq)}+ H^+_{(aq)}+ Cl^-_{(aq)} \rightarrow Na^+_{(aq)}+ Cl^-_{(aq)} + H_2O_{(L)} \qquad \Delta H_3~ =\end{align*}
The equation for calculating the heat absorbed by the water in a calorimeter is \begin{align*}q = mC \Delta t\end{align*}, where \begin{align*}q\end{align*} is the heat absorbed in Joules, \begin{align*}m\end{align*} is the mass of the water in the calorimeter, \begin{align*}C\end{align*} is the specific heat of water \begin{align*}(4.18 \ J/g \cdot ^\circ C)\end{align*}, and \begin{align*}\Delta t\end{align*} is the temperature change of the water due to the reaction.
Pre-Lab Questions (to be done before lab day):
- If the temperature of \begin{align*}500. \ g\end{align*} of water is raised by \begin{align*}3.5^\circ C\end{align*}, how many Joules of heat were absorbed?
- \begin{align*}100. \ mL\end{align*} of \begin{align*}0.100 \ M\end{align*} \begin{align*}NaCl\end{align*} contains \begin{align*}0.0100 \ moles\end{align*} of \begin{align*}NaCl\end{align*} \begin{align*}(moles = M \times L)\end{align*}. If \begin{align*}100. \ mL\end{align*} of \begin{align*}0.100 \ M\end{align*} \begin{align*}NaCl\end{align*} solution is diluted to \begin{align*}200. \ mL\end{align*}, how many moles of \begin{align*}NaCl\end{align*} will it contain?
Purpose:
- To physically measure and compare the amount of heat involved in three separate but related reactions.
- To provide experimental verification of Hess's Law.
Apparatus and Materials:
- 2 - large styrofoam cups (nested to provide better insulation)
- \begin{align*}100 \ mL\end{align*} graduated cylinder
- thermometer
- balance
- \begin{align*}150 \ mL\end{align*} beaker
- sodium hydroxide, \begin{align*}NaOH\end{align*} (solid)
- \begin{align*}0.50 \ M\end{align*} \begin{align*}NaOH\end{align*} solution (prepared the day before so it has time to cool)
- \begin{align*}0.50 \ M\end{align*} \begin{align*}HCl\end{align*} solution
- \begin{align*}0.25 \ M\end{align*} \begin{align*}HCl\end{align*} solution
- tap water
Safety Issues:
Hydrochloric acid and sodium hydroxide are corrosive. Avoid contact with skin. If any touches your skin, wash it off immediately. Solid sodium hydroxide is especially dangerous because it absorbs moisture rapidly from the air, forming an extremely concentrated solution. Avoid spilling this solid, and if a spill occurs, it must be cleaned up immediately. Ask your teacher for help. Be sure to close the lids of bottles of \begin{align*}NaOH\end{align*} securely immediately after using. A lab coat or apron and goggles are required for this lab.
Procedure:
Part One: The dissolving of solid \begin{align*}NaOH\end{align*} in water.
1. Put \begin{align*}200. \ mL\end{align*} of room temperature water into your nested styrofoam calorimeter. Stir carefully with a thermometer until a constant temperature is reached. Record this temperature.
2. Acquire about \begin{align*}2 \ grams\end{align*} of solid \begin{align*}NaOH\end{align*}, determine its exact mass and record the mass in your data table. It is necessary to perform this operation as quickly as possible to avoid too much water absorption by the \begin{align*}NaOH\end{align*}. (This water absorption not only alters its mass but begins the reaction before you are ready.)
3. Place the solid \begin{align*}NaOH\end{align*} into the water in your calorimeter. Stir gently with the thermometer until the solid is completely dissolved and record the highest temperature reached.
4. Discard the solution down the sink with excess water and rinse the cup and thermometer with water.
Part Two: The reaction of solid \begin{align*}NaOH\end{align*} with dilute \begin{align*}HCl\end{align*}.
5. Repeat steps 1 - 3 but replace the \begin{align*}200 \ mL\end{align*} of water with \begin{align*}200. \ mL\end{align*} of \begin{align*}0.25 \ M\end{align*} \begin{align*}HCl\end{align*}.
6. Again discard the solution and rinse your equipment with water.
Part Three: The reaction of \begin{align*}NaOH\end{align*} solution with hydrochloric acid solution.
7. Accurately measure \begin{align*}100. \ mL\end{align*} of \begin{align*}0.50 \ M\end{align*} \begin{align*}HCl\end{align*} solution and pour it into your calorimeter. Accurately measure \begin{align*}100. \ mL\end{align*} of \begin{align*}0.50 \ M\end{align*} \begin{align*}NaOH\end{align*} solution and place it in a \begin{align*}150 \ mL\end{align*} beaker. Measure and record the temperature of each solution. Be sure to rinse and dry your thermometer when switching solutions.
8. Add the sodium hydroxide solution to acid solution in your calorimeter. Stir the solution and record the highest temperature reached.
9. Discard solution and rinse equipment.
Data Table
Part One
Assuming the density of water to be \begin{align*}1.0 \ g/mL\end{align*}, mass of water = ___________ g
Mass of added solid \begin{align*}NaOH\end{align*} = ______________ g
Total mass of solution = ____________ g
Molar mass of \begin{align*}NaOH\end{align*} = ______________g
Initial temperature of water = _______________ g
Final temperature of solution = ____________ g
\begin{align*} \Delta t\end{align*} = ____________ \begin{align*}^\circ C\end{align*}
Part Two
Assuming the density of the \begin{align*}HCl\end{align*} solution to be \begin{align*}1.0 \ g/mL\end{align*}, mass of acid solution = ___________ g
Mass of \begin{align*}NaOH\end{align*} added = _____________ g
Total mass of solution = ___________ g
Initial temperature of \begin{align*}HCl\end{align*} solution = _____________\begin{align*}^\circ C\end{align*}
Final temperature of solution = _____________\begin{align*}^\circ C\end{align*}
\begin{align*} \Delta t\end{align*} = ____________ \begin{align*}^\circ C\end{align*}
Part Three
Assuming the density of both the \begin{align*}HCl\end{align*} and \begin{align*}NaOH\end{align*} solutions to be \begin{align*}1.0 \ g/mL\end{align*},
mass of \begin{align*}HCl\end{align*} solution = ___________ g
mass of \begin{align*}NaOH\end{align*} solution = __________ g
total mass of mixture = __________ g
Initial temperature of \begin{align*}HCl\end{align*} solution = _____________\begin{align*}^\circ C\end{align*}
Initial temperature of \begin{align*}NaOH\end{align*} solution = ____________\begin{align*}^\circ C\end{align*}
Average initial temperature = _____________\begin{align*}^\circ C\end{align*}
Final temperature of mixture = _____________\begin{align*}^\circ C\end{align*}
\begin{align*} \Delta t\end{align*} = ____________ \begin{align*}^\circ C\end{align*}
Calculations
Note: While the chemical properties of very dilute solutions can be quite different from the chemical properties of water, the physical properties of very dilute solutions are nearly the same as water. In this lab, the density and specific heat of the dilute solutions can be assumed to be the same as water.
1. Calculate the heat absorbed by the solution in the calorimeter in part one.
2. Calculate the moles of \begin{align*}NaOH\end{align*} dissolved in part one.
3. Calculate the heat released per mole of \begin{align*}NaOH\end{align*}.
4. Determine \begin{align*} \Delta H_1\end{align*}. (Keep in mind that exothermic reactions are assigned negative \begin{align*} \Delta H\end{align*} values.)
5. Calculate the heat absorbed by the solution in the calorimeter in part two.
6. Calculate the moles of \begin{align*}NaOH\end{align*} reacted in part two.
7. Calculate \begin{align*} \Delta H_2\end{align*}.
8. Calculate the heat absorbed by the solution in part three.
9. Calculate the number of moles of \begin{align*}NaOH\end{align*} reacted in part three.
10. Calculate \begin{align*} \Delta H_3\end{align*}.
11. Reactions 1 and 3 will add together to produce reaction 2. Show this is true. Therefore, according to Hess's Law, \begin{align*}\Delta H_1 + \Delta H_3\end{align*} should equal \begin{align*}\Delta H_2\end{align*}. Assuming \begin{align*} \Delta H_2\end{align*} to be the true value, calculate your percent error.
\begin{align*}\% \ error = \frac{true \ value-experimental \ value}{true \ value} \times 100 =\end{align*}
Post-Lab Questions
- Suppose you had used \begin{align*}4 \ grams\end{align*} of \begin{align*}NaOH\end{align*} instead of \begin{align*}2 \ grams\end{align*} in part one.
- How would the answer to calculation 1 have changed?
- How would the answer to calculation 2 have changed?
- How would the answer to calculation 3 have changed?