Topic 6: Energetics
6.1 Exothermic and Endothermic Reactions
6.1.1. Define the terms exothermic reaction, endothermic reaction and standard
enthalpy change of reaction (ΔHo).
Standard enthalpy change is heat transferred under standard conditions- pressure 101.3
kPa, temperature 298 K. Only ΔH can be measured, not H for the initial or final state of
a system.
Energy is conserved in chemical reactions. If stronger bonds form in the products than
are broken in the reactants, heat is released to the surroundings, and the reaction is
termed exothermic. If stronger bonds break than are formed, heat must be absorbed
from the surroundings, and the reaction is endothermic. Because strong bonds are more
apt to form than weak bonds, spontaneous exothermic reactions are common—for
example, the combustion of carbon-containing fuels with air to give CO2 and H2O,
both of which possess strong bonds. Spontaneous endothermic reactions, however, are
also well known; the dissolving of salt in water is one example.
Standard Enthalpy of Reaction is defined as the heat transferred under standard
conditions-pressure 101.3 kPa, temperature 298 K. However only ΔH can be
measured, not H for the initial or final state of a system. So, that is, the amount of
energy transferred either from the surroundings to the system or vice versa in a
reaction.
6.1.2. State the relationship between temperature change, enthalpy change and
whether a reaction is exothermic or endothermic.
Combustion of organic compounds are good examples of exothermic reactions.
Generally speaking, if the temperature around the system increases during and after a
reaction, then the reaction is exothermic because it releases energy in the form of heat
into the environment. On the other hand though, if the surroundings of a reaction
decreases in temperature, then the reaction is probably endothermic because it is
absorbing energy from the surroundings in the form of heat. This also applies to enthalpy
change, if a reaction absorbs energy, then it has a positive enthalpy change, if it releases
energy, it has a negative enthalpy change.
6.1.3. Deduce, from an enthalpy level diagram, the relative stabilities of reactants and
products and the sign of the enthalpy change for the reaction.
If the final state is more stable (lower on the enthalpy level diagram), this implies that
Hfinal< Hinitial and ΔH must be negative. Energy must be released in going to a more
stable state.
Generally speaking, substances are more stable if they are lower on an enthalpy level
diagram and less stable if they are higher on an enthalpy level diagram. So, if you follow
a reaction, and it starts out high on an energy level diagram and after the reaction the
products are lower, then the reaction has released energy in an effort to go from lower
stability reactants to higher stability products. It releases energy, so ΔH must be
negative.
6.1.4. Describe and explain the changes which take place at the molecular level in
chemical reactions.
Relate bond formation to the release of energy and bond breaking to the absorption of
energy.
At the molecular level, a chemical reaction primarily consists of the forming and
breaking of chemical bonds. When bonds break between the reactants, energy must be
absorbed in order to break those bonds. However, when bonds reform in creating the
products, energy is released into the environment. So it takes energy to break the original
bonds, but it is released again, and sometimes more then it took, in reforming the bonds.
6.1.5. Suggest suitable experimental procedures for measuring enthalpy changes of
reactions in aqueous solution.
Explore different reactions operating at constant pressure (open containers). Use of the
bomb calorimeter is not required.
Since when a reaction takes place in aqueous solution energy would either be released or
absorbed, this could be measured by measuring the temperature of the water. You want
the water to be heavily protected from external temperature changes. So, for example,
you could have five reactions and you want to find out which ones release more or absorb
more energy. You have them all take place in an aqueous solution under the same
pressure, and observe the temperature change. Whichever one releases more energy will
cause the temperature to increase the most, and whichever one absorbs more energy will
decrease the temperature the most.
6.2 Calculation of Enthalpy Changes
6.2.1. Calculate the heat change when the temperature of a pure substance is altered.
Students should be able to calculate the heat change for a substance given the mass,
specific heat and temperature change.
To do this, you use the equation change in energy (or heat)= mass*specific heat
capacity*Δt (change in temperature). Be sure to keep in mind how many moles are
reacting.
6.2.2. Explain that enthalpy changes of reaction relate to specific quantities of either
reactants or products.
Enthalpy changes are measured in joules (J) and are often quoted in kJ mol-1 of either a
reactant or a product.
Enthalpy changes are related to the number of moles in the reaction. If all the
coefficients are doubled, then the value of ΔH will be doubled. Also remember to pay
attention to limiting reagents though.
6.2.3. Analyze experimental data for enthalpy changes of reactions in aqueous
solution.
To do this, you use the equation E (energy or ΔH)= mass (of the water/aqueous solution)
* c(4.18 kJ kG-1 K-1) * ΔT (K). This value can then be calculated back to find the
enthalpy change for each mole of reactants.
6.2.4. Calculate the enthalpy change for a reaction in aqueous solution using
experimental data on temperature changes, quantities of reactants and mass of
solution.
Enthalpy change of an acid-base reaction could be investigated.
The results of a reaction will be a change in temperature. This can be converted into a
change in heat (or energy) by using the above equation and a known mass of water. This
can be used to calculate the ΔH for the amount of reactants present, and then this can be
used to calculate ΔH for a given number of moles.
6.3 Hess’s Law
6.3.1. Determine the enthalpy change of a reaction which is the sum of two or more
reactions with known enthalpy changes.
Use examples of simple two and three-step processes. Students should be able to
construct simple enthalpy cycles, but will not be required to state Hess’s law.
Hess’s law states that the total enthalpy change between given reactants and products is
the same regardless of any intermediate steps (or the reaction pathway.) To calculate,
you follow the following steps.
1. Reverse any reactions which are going the wrong way and invert the sign of their
ΔH values.
2. Divide or multiply the reactions until the intermediate products will cancel out
when the reactions are vertically added (always multiply/divide the ΔH value by
the same number.)
3. Vertically add them.
4. Divide or multiply the resulting reaction to the correct coefficients.
6.4 Bond Enthalpies
6.4.1. Define the term average bond enthalpy.
Bond enthalpies are quoted for the gaseous state and should be recognized as average
values obtained from a number of similar compounds. Cross reference with 11.2.6.
This is simply a different method of calculating the enthalpy of a reaction. ΔH=sum of
the energies required to break old bonds (positive signs) plus the sum of the energies
released in the formation of new bonds (negative signs). Keep in mind however that
bond enthalpies are always quoted from the gaseous state, and that the values you find for
certain bond enthalpies like a C-H bond in a table is just the average of bond enthalpies,
there are many different enthalpies that are somewhat variant. But basically a bond
enthalpy is how much energy it takes to break a bond or reform it. You just find what
bonds need to be broken then what bonds need to be reformed, and add up the numbers
for those bonds.
6.4.2. Calculate the enthalpy change of a reaction using bond enthalpies.
See above. It just takes practice. Say you want to calculate the enthalpy change of the
following reaction:
H2 + F2  2HF.
You would first find out how much energy it takes to break the H2 bonds and F2 bonds,
sum them up, and subtract from them the amount of energy it takes to reform new bonds.
6.5 Entropy
6.5.1. State and explain the factors which increase the disorder (entropy) in a system.
An increase in disorder can result from the mixing of different types of particles, change
of state (increased distance between particles), increased movement of particles or
increased number of particles. An increase in the number of particles in the gaseous
state usually has a greater influence than any other possible factor.
Entropy can be viewed as a measure of randomness or disorder. The natural progression
of things is from order to disorder, from lower entropy to higher entropy. Things that
tend to increase entropy are the mixing of different types of particles, the change of state
of those particles (different states have different amounts of distance between them, the
more distance the more entropy), increased movement of particles or increased number of
particles. In the gaseous form, and increased number of particles usually has the most
effect on entropy.
6.5.2. Predict whether the entropy change (ΔS) for a given reaction or process would be
positive or negative.
From a given equation, identify a single factor which affects the value of ΔS and predict
the sign of ΔS.
Predict the sign of ΔS for a reaction based on the above factors. ΔS is positive when
entropy increases (more disorder) and negative when entropy decreases (less disorder).
6.6 Spontaneity
6.6.1. Define standard free energy change of reaction (ΔGo).
A thermodynamic function equal to the enthalpy (H) minus the product of the entropy (S)
and the Kelvin temperature (T); G=H – TS. Under certain conditions the change in free
energy for a process is equal to the maximum useful work.
6.6.2. State whether a reaction or process will be spontaneous by using the sign of ΔGo.
When ΔGo for a reaction is negative, then the reaction is spontaneous. When it is
positive, then the reaction is not spontaneous.
6.6.3. State and predict the effect of a change in temperature on the spontaneity of a
reaction, given a standard entropy and enthalpy changes.
ΔG= ΔH-TΔS. Just figure how the temperature effects each individual circumstance
using that equation. For example if entropy is negative and enthalpy is positive, then a
higher temperature would only increase the unsponteneity of the reaction, but if enthalpy
is positive and entropy is positive, then a higher temperature would make the reaction
more spontaneous. It really depends on the case.