Chemistry 30
Thermodynamics
Two of the Laws of Thermodynamics (simplified)
The 1st Law: Energy can be neither created nor destroyed. It can only change forms.
In any process in an isolated system, the total energy remains the same. (The energy of the universe is constant.)
The 2nd Law: "heat can spontaneously flow from a higher-temperature region to a
Thermodynamics is the study of the energy changes of chemical and physical process. Put in the simplest possible terms, this unit concerns chemical potential kinetic energy, and chemical potential energy.
Chemical Kinetic Energy, Ek
For simple molecules EkAVE = 3/2 RT (where R = 8.314 J/K.mol, and T is the temperature in Kelvin)
No matter the states/phases of different samples of matter, if the samples are all at the same temperature, the molecules have exactly the same kinetic energy.
Molecules above absolute zero always have kinetic energy. That's why they are incessantly speeding through space and hitting each other and rotating and vibrating in a gas or liquid (… see opposite). Even in solids where they can't move around, they are always vibrating. Molecules tend to spread out their energy by moving and rotating/vibrating in as many different ways as they can, given the constraints of the state they are in.
For molecules/atoms in the solid state
Ek Total =
Ek (vibration)
+ Ek (rotation) + Ek (translation) (mostly vibrational Ek) In the solid state rotation and translation are restricted).When food containing liquid water is microwaved the water molecules end up rotating at a much faster rate. (The microwave radiation has the same frequency/phase as water molecule rotation) The food containing this water warms up. Molecules with more Ek are hotter. When ice is microwaved it does not warm up appreciably because the water molecules in the ice (solid state) are so hydrogen bonded that they are essentially not rotating at all. Obviously, their rate of rotation cannot be increased! You may have noticed that food microwaved from the frozen state is often still rock hard/frozen inside! For molecules/atoms in the gas state
kJ kg.°C
Calculating Kinetic Energy Losses and Gains.
The change in Ek of a warming or cooling can be calculated using: q = m c t Where q = the quantity of kinetic energy, in kJ, absorbed or released
m = sample mass, in kg
t = temperature increase/decrease/change t, in °C, and
c = the specific heat capacity of the sample of matter (…essentially the quantity of Ek ( kJ or J) needed to warm a unit mass of matter (kg or g) by 1°C). The change in temperature of a substance, when it loses or gains a certain quantity of heat energy, is inversely proportional to the heat capacity of that substance. Or to put it another way, a substance with a high heat capacity absorbs more heat energy for a particular temperature change than the same mass of another substance with a lower heat capacity.
Homogeneous mixtures of substances, air for example, often exhibit the weighted average heat capacity of their components. The average heat capacity of pure water between 0-100 °C, 4.19 kJ.kg.°C, is so large that aqueous solutions less than 2 mol/L, to a close approximation, have essentially the same heat capacity as that of water. Knowing the mass of water in a thermodynamic experiment is always the best practice, but in an experiment involving aqueous solutions the density of water is such that every millilitre of solution contains close to one gram of water. So 100 mL of 1.0 mol/L HCl(aq) has 100 g of water within it.
A unit: is fully equivalent to J
g.°C
(and in turn to J
g.K
for IBO). Specific heat capacities in textbooks, etc, are always quoted as J
g.°C
.
In practice, even small temperature changes for the quantities of substance employed in high school laboratories often involve kilojoule quantities of energy. So, it is good practice when calculating Ek values to use substance masses in kilograms. (To a good approximation, 1 mL of water has a mass of 1g)
Chemistry 30
1. Equal quantities of olive oil and water in separate, identical, beakers are placed on opposite sides of a hot plate. This way, equal quantities of heat energy are
transferred into each liquid. When the water boils will the oil have a higher, a lower, or the same temperature as the water?
3. Assuming a specific heat capacity the same as pure water, calculate the heat energy that must be removed from a 250 mL soft drink from 22 °C to 5 °C. (18 kJ)
4. What mass of water absorbs 1.00 MJ of heat energy, when its temperature rises from 22 °C to 100 °C? (3.1 kg)
5. What mass of aluminum absorbs 1.00 MJ of heat energy, when its temperature rises from 22 °C to 100 °C? (14 kg)
6. What mass of copper absorbs 1.00 MJ of heat energy, when its temperature rises from 22 °C to 100 °C? (33 kg)
8. Calculated the total energy absorbed when 1000 g of water, in a 250 g copper pot, are warmed from 10 °C to 100 °C. (0.39 MJ)
9. Equal 100 g masses of Cu(s), Al(s), and Fe(s) (cFe = 0.444 J/g°C) at 300 °C are placed into separate identical calorimeters each containing 200g of water at 22.0°C. Which calorimeter will have the highest final temperature? Explain.
Chemical Potential Energy, Ep
Potential energy is absorbed to separate atoms in an electrostatic field, and released when atoms come together in that same field. Or, to put it another way, energy is absorbed to break bonds and energy is released when bonds are broken.
Covalent bonding - overview Electrons have a negative charge and the nuclei they surround have a positive charge.
When two atoms approach one another the electron clouds may overlap. When this happens, under certain circumstances, the bonding electrons that are between two nuclei can be attracted to both nuclei, holding them together. This would be a single bond.
When the atoms form a bond they become lower in energy - the system is more stable.
The energy saved by moving to a more stable situation is released as heat. Thus, bond formation is always exothermic, i.e. heat energy is released.
Conversely, in order to break a chemical bond energy must be put in - it is an endothermic process. (Ein to break, Eout to make – bonds)
Obvious examples of bond formation abound. When any liquid solidifies, or any gas condenses, energy is released as intermolecular bonds form. Ice forms as heat energy is removed from liquid water at 0 °C. Conversely, as water/perspiration on your
skin/clothing evaporates it has a pronounced cooling effect as energy is absorbed to break hydrogen bonds between the water molecules in the liquid state.
To find the energy of something that is heating up or cooling down(TEMP CHANGE), which is a kinetic energy change always use q = m c t. Where is the energy change, enthalpy change, either absorbed or released from or to a system in joules(J).
When there is a potential energy change we use the formula qnrxn. The enthalpy
change is equal to the moles of species reacting multiplied by the molar enthalpy of reaction. is measured in kJ and molar enthalpy is in kJ/mol.These questions are very easy to spot…there are enthalpy changes with NO TEMPERATURE CHANGE of the system. They can be for example
Chemical reactions ( formation, decomposition, single replacement, precipitation and combustions). All the reactions you did in grade 10.
o Dissolving, other chemical reactions and phase changes are some other ones ( I have not seen phase changes on the diploma for over 10 years) These are basically and potential energy changes due to bond breaking and reforming in a chemical system.
Think of the energy released when burning fully 1 match. If you burn 2 matches you will release twice as much energy. Hence the heat change is recorder per mole, sometimes per gram, of a substance reacting.
1. If the heat of formation of water gas is -241.8 kJ/mol , page 5 of your data
booklet. How much energy is released when 2 moles of water are formed from its elements? ( -483.6 kJ)
2. What is the heat of combustion of ridiculous acid if 0.5 moles of it were
combusted and was found to release 23.4 kJ? ( -46.8 kJ/mol). Why is the answer negative?
4. The heat of dissolving of sodium chloride is +3.87 kJ/mol. How many moles of salt need to be dissolved to absorb 57.0 kJ of energy. ( 14.7 mol of NaCl)
5. What is the heat of solvation of ammonium nitrate if the dissolving of 120.0 grams of it caused the water it was dissolved in to lose 38.5 kJ of energy? (+25.7 kJ/mol)
6. Here are some heats of combustion:
Name Formula combustion
Methane CH4 (g) -802.3 kJ/mol
Ethane C2H6 (g) -1,437.2 kJ/mol
Propane C3H8 (g) -2,044.2 kJ/mol
Calorimetry is the quantitative measurement of the heat lost/gained during a
chemical/physical process. A Calorimeter is a device for gathering measurements to calculate heat of a reaction. In a perfectly efficient system, and in accordance with the 1st and 2nd Laws of Thermodynamics, the energy lost by an exothermic reaction to its surroundings, and therefore the energy gained by the surroundings, are equivalent. Similarly, the energy gained by an endothermic reaction from its surroundings is equivalent to the energy lost by the surroundings. Most calorimeters surround a chemical reaction with water. So Hgain+Hloss= 0 which can be rearranged into
H
gain
= -H
loss
If the surroundings of a reaction only warm up or cool down then:
Ep(reaction) = Ek(surroundings)
Since Ep = n rxnH° and Ek = m c t
Then n rxnH° = m c t
in other words the enthalpy change of a reaction, at constant pressure, is equivalent to the kinetic energy change of its surroundings.
Simple Cup Calorimetry
We will use a simple Styrofoam cup calorimeter in the high school laboratory, which, by its construction, is necessarily at constant pressure:
Assumptions:
1. Styrofoam provides perfect insulation 2. Only the water in
the calorimeter warms/cools. 3. Vsolution = VH2O 4. csolution = cH2O
Thus: n rxnH° = m c t
Where n = moles of reactant;
rxnH° = the molar enthalpy of reaction for that reactant
1. When 50.0 ml of 1.0 mol/L HCl(aq) and 50.0 mL of 1.0 mol/L KOH(aq) are mixed the temperature of the product solution is 6.80 °C higher than the reactant initial temperatures. Calculate the molar enthalpy of neutralization of the acid. (What is the molar enthalpy of neutralization of the base? Explain.) (-57 kJ/mol)
2. Some 8.00 g of fresh sodium hydroxide are added to 100 g of deionized water in a Styrofoam cup calorimeter, initial temperature 20.6 °C. The reactant is gently swirled and eventually it reaches a maximum temperature of 40.4 °C. What is the molar enthalpy of solution for the salt? (-41.5 kJ/mol)
3. When samples of sodium nitrate are added to water in a coffee cup calorimeter the following data were obtained
Measurements for dissolving NaNO3(s) in Water
Quantity Trial #1 Trial #2
Mass of Cups (g) 6.74 6.78
Mass of Cups & water (g) 105.80g 106.93 Mass of NaNO3(s) (g) 16.00 16.05 Initial calorimeter temperature (°C) 20.5 20.5 Final calorimeter temperature (°C) 12.9 12.7
4. When a 10.0 g sample of pure sulfuric acid is added to 100 mL of water in a
calorimeter. The initial temperature of the water is 20.0 °C. The highest temperature reached by the solution is 31.2 °C.
a. What is the energy change of this reaction? (4.69 kJ)
b. What is the molar enthalpy of solution of the acid? (-46 kJ/mol)
5. When samples of magnesium strip were dropped into nitric acid in a simple cup calorimeter the following data was obtained
Measurements for the reaction of Mg(s) and 1.00 mol/L HNO3(aq)
Quantity Trial #1 Trial #2
Mass of Mg (s) 0.21 0.22
Volume of HNO3(aq) (mL) 50.0 50.5 Calorimeter tinitial (°C) 20.5 20.5 Calorimeter tfinal (°C) 39.1 39.6
Calculate the enthalpy of reaction of magnesium for each trial, and an average enthalpy for the reaction. (-0.45 MJ/mol)
6. The enthalpy of solution of potassium nitrate is +34.89 kJ/mol. Some 0.100 mol of KNO3(s) is dissolved in 100 g of water in a simple cup calorimeter, initial
7. A student performs a calorimetry experiment to determine the molar enthalpy of combustion of ethanol, as represented by the equation below.
C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)
The burning ethanol, in a small spirit burner, heated a quantity of water in a pop can. The data for one trial collected during the experiment are recorded below.
Combustion Calorimetry Experiment Data
Mass of aluminium pop can 10.25 g
Mass of water 240.00 g
Final mass of ethanol burner 152.45 g Initial mass of ethanol burner 156.34 g Final temperature of water and can 80.0 °C Initial temperature of water and can 20.0 °C
Calculate the enthalpy of combustion of C2H5OH(l). (-721 kJ/mol)
More rigorous experiments give the value for the molar enthalpy of combustion of ethanol as –1235.6 kJ/mol . Calculate a % error for the experiment. Suggest two likely reasons for the difference in the two combustion enthalpies.
8. The calculated enthalpy of neutralization of any strong base by equal moles of
Bomb Calorimetry
Bomb calorimeters are so called because they have a disquieting habit of exploding while being pressurized with oxygen for a combustion reaction. Solid samples are pelletized for ease of handling. The following diagram depicts a constant volume or 'bomb' calorimeter:
The stirrer in a bomb calorimeter ensures that any heat released or absorbed by a reaction within the bomb is distributed evenly throughout the apparatus.
For a reaction in a bomb calorimeter:
n rxnH° = (m c t)H2O + (m c t)b + (m c t)st + (m c t)th + (m c t)con
(b = bomb, st = stirrer, th = thermometer, con = containers) Which reduces to:
n rxnH° = (m H2Oc H2O + m bc b + m stc st + m thc th + m conc con) t
Which further simplifies to: n rxnH° = C t
- since all the masses and specific heat capacities of the components are constant and where C = m H2Oc H2O + m bc b + m stc st + m thc th + m conc con. C, the heat capacity of
1. The heat of combustion of benzoic acid, C6H5COOH(s), is -26.38 kJ/g. When a 1.200 g sample of benzoic acid burns in a bomb calorimeter it causes the
calorimeter temperature to rise by 3.38 ºC. What is the heat capacity of the
calorimeter? (9.37 kJ/ºC)
2. Nitromethane, CH3NO2(l) burns to produce nitrogen and carbon dioxide gases and liquid water. Burning 2.00 g of Nitromethane in a bomb calorimeter, heat capacity 7.794 kJ/ºC, raises the apparatus temperature from 25.00 ºC to 27.98 ºC. Calculate the molar enthalpy of combustion of nitromethane. (-709 kJ/mol)
3. A reaction within a calorimeter absorbs 13.5 kJ of heat energy. The heat capacity of the calorimer is 19.57 kJ/ ºC. If the final temperature of the calorimeter was 24.31 ºC what was its initial temperature? (25.00 ºC)
4. When a 1.100 g sample of benzoquinone C6H4O2(s), burns in a calorimeter with a heat capacity of 3.930 kJ/ºC it causes the calorimeter temperature to rise from 20.02 ºC to 27.13 ºC. What is the molar enthalpy of combustion of
benzoquinone? (-2.75 × 103 kJ/mol)
Communicating/Interpreting Enthalpy Changes
1. Molar enthalpy of process, may be standard, (kJ/mol) rxnH° = ± X kJ/mol The molar enthalpy of formation of
aluminum oxide is communicated as fH° (Al2O3(s)) = –1675.7 kJ/mol The molar enthalpy of formation of
nitrogen monoxide is communicated as fH° (NO(g)) = +91.3 kJ/mol
2. The enthalpy of the reaction as written, (kJ): Reactants Products fH° = ± X kJ The formation of Al2O3(s) is 2 Al(s) + 3/2 O2(g) Al2O3(s) fH° = –1675.7 kJ The formation of NO(g) is ½ N2(g) + ½ O2(g) NO(g) fH° = +91.3 kJ
3. The enthalpy change as a reactant or product, (kJ): Endothermic reaction : Reactants + X kJ Products
Exothermic reaction : Reactants Products X kJ
forming Al2O3(s) is 2 Al(s) + 3/2 O2(g) Al2O3(s) + 1675.7 kJ forming NO(g) is ½ N2(g) + ½ O2(g) + 91.3 kJ NO(g)
3. The enthalpy graph:
Endothermic reaction Exothermic reaction
e.g. the formation of nitrogen monoxide (the products have higher total
enthalpy than the reactants) Label the reactants and products plateaux.
1. Communicate each of the following process in the four ways shown.
a. When a mole of magnesium sulfate forms, 1284.9 kJ of energy is released
b. The formation of zinc sulfide.
d. When a mole of potassium chlorate forms, 393.7 kJ of energy is released.
e. The simple decomposition of sucrose.
What is the origin/destination of reaction energy?
By definition the enthalpy of formation of any element in its standard state is 0 kJ/mol.
The thermodynamic stability of a compound is greater when it has a more
exothermic molar enthalpy of formation. For example CO2(g), fH° = -393.5 kJ/mol, is more thermodynamically stable than CO(g), fH° = -110.5 kJ/mol.
When methane burns (see below)
Chemical potential energy is absorbed when bonds in the reactants (CH4 and two O2) are broken. (Call this absorbed energy Ein)
Chemical potential energy is released as bonds form in the products (CO2 and two H2O) (Call this released energy Eout)
Overall, this reaction releases energy to its surroundings (it is exothermic). Obviously, the energy released must be greater than the energy absorbed! In other words, | Eout| > | Ein|. It follows therefore that for an endothermic reaction,
| Ein | > | Eout|.
energies exceed the activation energy is
Ea RT
e : where Ea = the activation energy in kJ/mol. R = 8.314 J/K.mol and T = the Kelvin temperature. On an energy profile graph the
activation energy is always represented as a hilly barrier between the reactants and the products
The activation energy of the forward reaction is the difference between the enthalpies of points II and I in the diagram shown here. (The activation energy of the reverse reaction is the difference between the energies of points II and III.) Generally speaking, the faster reaction has the lower Ea.
A catalyst speeds up a chemical reaction by providing an alternate reaction pathway that, overall, has faster steps. The slowest step has a lower Ea than the uncatalyzed reaction. A catalyst does not change a reaction’s H, see below.
A catalyst causes a reaction to lose or gain heat energy to to/from its surroundings at a faster rate. Notice that it decreases the forward and reverse activation energies
Questions.
1. (a) State the molar enthalpy of formation of hydrogen iodide gas. (+26.5 kJ/mol)
(b) What is the enthalpy change of the reaction illustrated below. (+53.0 kJ) H—H + I—I H—I + H—I rxnH° = ______?
Bonds broken Bonds formed
(c) How does the energy of breaking the bonds in the reactant elements compare to the energy of forming the product bonds?
2. Rank the chemicals species benzene, aluminum oxide, phosphorus, and methanol in terms of decreasing thermodynamic stability. Explain your ranking.
3. Rank the oxides: SnO(s), SnO2(s), HgO(s). MgO(s), and ZnO(s) in terms of the energy required, per mole of metal, needed to decompose each to their elements.
How does this ranking compare to the thermodynamic stability of these compounds?
Hess’s Law
The enthalpy change of a net reaction is the numeric sum of the enthalpies of all the reactions that were combined to make the net reaction.
This law states that the energy change for any chemical or physical process is independent of the pathway or number of steps required to complete the process, provided that the final and initial reaction conditions are the same. In other words, the enthalpy difference of a multistep process is simply the difference in enthalpies of the initial reactants and the final products.
Hess's law allows the enthalpy change (ΔH) for a reaction to be calculated even when it cannot be measured directly. This is usually accomplished by performing arithmetic operations on chemical equations of known ΔH values, doing the same with those
reactions’ enthalpies, and then combining the altered equations to get the net reaction and enthalpy desired. A common use of Hess’s law is to calculate standard molar enthalpies of formation of compounds much like those in your data booklet.
Example: Use the following reaction equations and thermodynamic data to calculate a molar enthalpy of formation for propene, C3H6(g)
1. C3H6(g) + 9/2 O2(g) 3 CO2(g) + 3 H2O(l) cH° = –2057.8 kJ 2. C(s) + O2(g) CO2(g) fH° = –393.5 kJ 3. H2(g) + 1/2 O2(g) H2O(l) fH° = –285.8 kJ
The goal of this problem is to manipulate the equations and enthalpies given to arrive at the equation 3 C(s) + 3 H2(g) C3H6(g) …fH° = ? kJ. In most cases if we alter (use multiples/reversals of) the given equations on the basis of looking after the compounds, the required quantity of elemental components will, in most cases, be obtained.
Reversing #1 gives C3H6(g) as a product and the reaction enthalpy becomes +2057.8 kJ.
Tripling #2 gives 3 CO2(g) as a product to cancel the CO2s formed upon reversing #1 and the reaction enthalpy becomes –1180.5 kJ.
Tripling #3 gives 3 H2O(l) as a product to cancel the H2Os formed upon reversing #1 and the reaction enthalpy becomes –857.4 kJ.
Now we have
1. 3 CO2(g) + 3 H2O(l) C3H6(g) + 9/2 O2(g) rxnH° = +2057.8 kJ 2. 3 C(s) + 3 O2(g) 3 CO2(g) + 3 H2O(l) fH° = –1180.5 kJ 3. 3 H2(g) + 3/2 O2(g) 3 H2O(l) fH° = –857.4 kJ Net: 3 C(s) + 3 H2(g) C3H6(g) fH° = +19.9 kJ When a mole of propene forms some 19.9 kJ of heat energy are absorbed. It is important to be able to calculate this, and other, formation enthalpies. Tabulated formation
Questions
1. Use the following equations 1/
2 N2(g) + 3/2 H2(g) NH3(g) ∆fH˚ = -74.6 kJ N2(g) 2 N(g) ∆dH˚ = +945.8 kJ H2(g) 2 H(g) ∆dH˚ = +435.2 kJ
to calculate the enthalpy an for the reaction that forms three moles of N-H bonds in ammonia
N(g) + 3 H(g) NH3(g) ∆˚netH = ? (-1200.3 kJ)
2. One way to produce hydrogen chloride, HCl(g), in the laboratory is to drip concentrated sulfuric acid on potassium chloride according to the equation 2 KCl(s) + H2SO4(l) 2 HCl(g) + K2SO4(s) ∆˚netH = ?
Use the following chemical reaction equations
2 KOH(s) + H2SO4(l) 2 H2O(l) + K2SO4(aq) ∆˚rH = -342.4 kJ HCl(g) + KOH(s) KCl(s) + H2O(l) ∆˚rH = -203.6 kJ K2SO4(s) K2SO4(aq) ∆˚sH = +23.8 kJ -to calculate the enthalpy for the reaction potassium chloride and sulfuric acid. (+41.0 kJ)
3. From the following reactions and their enthalpies
O2(g) 2 O(g) ∆˚rH = +494.0 kJ 3 O2(g) 2 O3(g) ∆˚rH = +285.3 kJ
- calculate an enthalpy for the reaction represented by the following equation.
4. Calculate the molar enthalpy of formation of anthracene, C14H10(s), from the following reactions and their enthalpies (+123.0 kJ/mol)
2 H2(g) + O2(g) 2 H2O(l) ∆˚fH = –571.6 kJ C(s) + O2(g) CO2(g) ∆˚fH = –393.5 kJ C14H10(s) + 33/2 O2(g) 14 CO2(g) + 5 H2O(l) ∆˚cH = –7061.0 kJ
5. a) Write an equation for the formation of calcium carbide, CaC2(s).
b) Use the following reaction equations and enthalpies to determine a molar enthalpy of formation for calcium carbide.
CaO(s) + 3 C(s) CaC2(s) + CO(g) ∆˚rH = +464.8 kJ 2 C(s) + O2(g) 2 CO(g) ∆˚fH = -221.0 kJ 2 Ca(s) + O2(g) 2 CaO(s) ∆˚fH = -1269.8 kJ - calculate a molar enthalpy of formation of calcium carbide. (-59.6kJ/mol)
6. Manipulate the following reactions and their enthalpies
2 H2(g) + O2(g) H2O(g) ∆˚fH = –571.6 kJ C(s) + O2(g) CO2(g) ∆˚fH = –393.5 kJ 2 C(s) + O2(g) 2 CO(g) ∆˚fH = –221.0 kJ C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) ∆˚cH = –2043.9 kJ
To calculate an enthalpy for the incomplete combustion of propane according to the equation C3H8(g) + 7/2 O2(g) 3 CO(g) + 4 H2O(g) ∆˚cH = ? (-1194.9 kJ)
7. A catalyst has been developed for the reaction of octane to form carbon dioxide and hydrogen gas, for a fuel cell, as follows
C8H18(l) + 8 O2(g) 8 CO2(g) + 9 H2(g) Use the following reactions and their enthalpies
H2(g) + 1/2 O2(g) H2O(g) ∆˚fH = –241.8 kJ C(s) + O2(g) CO2(g) ∆˚fH = –393.5 kJ C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(g) ∆˚cH = –5074.1 kJ
Calculating Reaction Enthalpies using Tabulated fH° values.
For any balanced chemical equation, for which the formation enthalpies of the reactants and products are available, it is possible to calculate the net enthalpy change using the following equation:
netH° = n fH°(products) - n fH°(reactants)
Example: What is the enthalpy change for the complete combustion of propane to give gaseous products?
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) cH° = ?
netH° = n fH°(products) - n fH°(reactants)
2 2 3 8
-393.5 -241.8 -103.8
3 1
- 2043.9
(
{
}
{
}) - (
{
})
net
kJ
H mol mol mol
molCO molH O molC H
kJ 4
When a mole of propane burns some 2043.9 kJ of heat energy are released. 1. Calculate the enthalpy of combustion for one mole of each of the following fuels
when they burn to give gaseous products: a. Methane (-802.5 kJ)
b. Ethanol (-1234.8 kJ)
c. Methanol (-637.9 kJ)
d. Octane (-5074.1 kJ)
2. Calculate the enthalpy changes of the following reactions: a. Photosynthesis of glucose (+2802.5 kJ)
b. Cell respiration of glucose (-2802.5 kJ)
c. Yeast fermentation of glucose according to the unbalanced equation: ___C6H12O6(s) ___CO2(g) + ___ C2H5OH(l) (-68.9kJ)
d. Fermentation of glucose by methanogenic bacteria according to the
unbalanced equation:
___C6H12O6(s) ___CO2(g) + ___ CH4(g) (-131.0 kJ)
e. Addition of water to ethene entirely in the gas phase. (-88.2 kJ)
f. The balanced single replacement reaction of iron and aluminum oxide. (+851.5 kJ)
g. Production of hydrogen according to the reaction equation:
CH4(g) + H2O (g) CO(g) + 3 H2(g) (+205.9 kJ)
Thermodynamic data can be provided in a form other than a table – see below Use the following information to answer the next question
3. Calculate the enthalpy changes when
a. 0.530 mol of VCl4(l) decomposes to form VCl2(s) and Cl2(g). (+62.2 kJ)
b. 1.00 mol of Cl2(l) reacts with sufficient VCl2(s) to form VCl3(l). (-128.7 kJ)
Calculating Product/Reactant fH° from Reaction Enthalpies, netH°.
We know that: netH° = n fH°(products) - n fH°(reactants)
Upon rearranging this equation we obtain two other useful equations:
n fH°(reactants) = n fH°(products) - netH°
n fH°(products) = n fH°(reactants) + netH°
Example 1. The standard molar enthalpy of combustion of acetophenone,
CH3C(O)C6H5(g) is –4140 kJ/mol. What is the standard molar enthalpy of formation of acetophenone?
First we must write a balanced combustion for the acetophenone.
CH3C(O)C6H5(g) + 2 O2(g) 8 CO2(g) + 4 H2O(l) cH° = –4140 kJ Clearly: n fH°(reactants) = n fH°(products) - netH°
2 2
3 6 5
3 6 5
-393.5 -285.8
1 [ ] 8 4 -4140
1 [ ( ) ] -151
[ ( ) ] -151 /
(
{
}
{
}) - (
)
f f f kJ kJ
mol H R mol mol kJ
molCO molH O
mol H CH C O C H kJ
H CH C O C H kJ mol
The standard molar enthalpy of formation of acetophenone is -151 kJ/mol. (For a standard molar enthalpy of combustion liquid water is produced in the confines of a bomb calorimeter surrounded by cold water.)
Occasionally, the molar enthalpy of reaction of the chemical in question must be determined before proceeding – this is a stoichiometry problem.
Example 2. If 47.80 kJ of heat energy are released when 1.000 g of 2,2,4 trimethylpentane, C8H18(l), burns to give products that include liquid water. What are the molar enthalpy of combustion, and formation, of 2,2,4 trimethylpentane?
8 18
1 114.26
-47.80 - 5462 /
1.000
cHo kJ g kJ mol
g mol C H
C8H18(l) + 25/2 O2(g) 8 CO2(g) + 9 H2O(l) cH° = –5462 kJ
8 18
-393.5 -285.8
1 [ ]
(
8{
}
9{
}) - (
-5462)
f
kJ kJ
mol H C H mol mol kJ
1. Calculate the standard molar heat/enthalpy of formation of substance X given the
following information (-120.0 kJ)
Substance fH° (kJ/mol)
M –22.5
Q +78.3
R –54.8
Given: M + 3Q 2 R + 2X rxnH° = –562.0 kJ
2. What is the standard molar enthalpy of formation of nitromethane, CH3NO2(l), if its standard molar enthalpy of combustion is –710 kJ/mol? ( The products are carbon dioxide and nitrogen gases, and liquid water.) (-112 kJ/mol)
3. Isobutane, IUPAC name methylpropane, is a common component of camping stove fuel that performs well at low temperature. Calculate a standard molar enthalpy of formation for methylpropane, CH3CH(CH3)CH3(g), if its standard molar enthalpy of combustion is –2869.0 kJ/mol? (+86.0 kJ/mol)
4. If the standard molar enthalpy of combustion of cyclopropane, C3H6(g), is
5. If 26.38 kJ of heat energy is transferred when 1000 mg of ethanal, CH3C(O)H(l), burns in a bomb calorimeter. Write the balanced chemical equation for the combustion of ethanal and calculate a molar enthalpy of combustion for ethanal.
(-1162 kJ/mol)
Use the enthalpy of combustion calculated above to estimate a molar enthalpy of
formation for ethanal. (-108.6 kJ/mol)
6. When dimethyl ether, (CH3)2O(g), burns in a diesel engine the following reaction occurs:
(CH3)2O(g) + 3 O2(g) 2 CO2(g) + 3 H2O(g) cH = –1086.5 kJ Calculate a molar enthalpy of formation for dimethyl ether. (-425.9 kJ/mol)
7. A component of most gasolines today is the octane booster MTBE, 2-methoxy-2-methylpropane and formula (CH3)3COCH3(l). When it burns in an engine the following reaction occurs:
Practical/Real Thermodynamics.
The most common source of the energy that fuels most nations’ economies comes from burning fossil fuels (coal, natural, petroleum). In practical terms we do not distribute or buy fuels in molar amounts, nor do we normally associate negative signs with the exothermic reactions we employ. While many fuels are sold by the tonne/barrel/litre, natural gas, mostly methane, is sold per gigajoule of energy it produces upon combustion. Real thermodynamics revolves around calculating:
The loss/gain in energy for any reactant or product, in any quantities (stoichiometry). Specific fuel enthalpies, usually in MJ/kg, are important.
The quantitative aspects of heating apparatus such as calculation of mass of water heated, mass of fuel consumed, temperature change of water.
Comparing conventional and unconventional fuels (e.g. biofuels).
Example1. Natural Gas, mostly methane, is sold by the Gigajoule (109 J). Given that the molar enthalpy of combustion of methane is –802.5 kJ/mol what is the mass of methane that must burn to produce 1.00 GJ of heat energy?
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) cH° = –802.5 kJ
4
4
6 4
4
16.05
1.00 10 2.00 10 ( 20.0 )
802.5
mCH
mol CH g
kJ g or kg
kJ mol CH
Evidently 20.0 kg of methane burn to produce 1.00 GJ of heat energy.
Example 2. The molar enthalpy of combustion of methane to give gaseous products is – 802.5 kJ/mol. What quantity of energy is produced per 1.00 kg of methane?
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) cH° = –802.5 kJ
4
3 4
4
802.5
1.00 10 5.00 10 ( 50.0 )
16.05
c
mol CH kJ
H g kJ or MJ
g mol CH
1. Calculate the energy released on burning the masses of the following auto fuels.
a. 1.00 kg of octane. (44.4 MJ)
b. 1.00 kg of ethanol. (26.8 MJ)
c. 1.00 kg of methyl oleate, C17H33COOCH3(l), fH° = –724.64 kJ/mol. (37.4 MJ)
d. 1.00 kg of dimethyl ether, (CH3)2O(g), fH° = –184.1 kJ/mol. (28.8 MJ)
3. Estimate the amount of methanol that must burn in a spirit burner to raise the temperature of 250.0 g of water from 20.0 °C to 35.0 °C? Will your estimate be higher or lower than the amount actually required in the laboratory? Explain.
(0.789 g, will take more – heat losses)
4. The metabolism of sucrose can be represented by the following equation. C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(g) ΔHº = –5639.7 kJ Calculate the energy transferred when 6.00 g cube of sucrose is metabolized. (98.8 kJ)
5. In disposable lighters butane combustion is represented by the following equation. 2 C4H10(g) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) ΔHº = –5314.6 kJ
The enthalpy change for the combustion of 1.00 g of butane gas is __________ kJ. (45.7 kJ)
7. A student uses an insulated aluminium can to determine the molar enthalpy of solution for solid potassium nitrate. The student assumes that the can neither gains nor loses much heat during the experiment; and that the specific heat capacity for the final solution is the same as that of water. The data were collected and recorded in the
following table. (+33.1 kJ/mol)
Observation Quantity
Mass of aluminum can 25.45 g
Mass of aluminum can and water 155.45 g Mass of potassium nitrate 10.1 g Initial temperature of can and water 23.46 ºC Final temperature of can and contents 17.14 ºC
Using the tabulated data, calculate the molar enthalpy of solution of potassium nitrate. (+33.1 kJ/mol)
8. The formation reaction of sodium bicarbonate can be represented as follows. Na(s) + 1/
2 H2(g) + C(s) + 3/2 O2(g) → NaHCO3(s) + 947.7 kJ Calculate the enthalpy change when 9.60 g of oxidizing agent are consumed. (-190kJ)
9. The roasting of zinc sulfide is represented by the unbalanced equation: ___ ZnS(s) + ____ O2(g) → ___ ZnO(s) + ____ SO2(g) Calculate the enthalpy change of the reaction balanced with whole number coefficients. (-882.6 kJ)
Calculate the energy transferred when 1.00 Mg of zinc sulfide are roasted. (4.52 GJ)
10. Calculate the energy released for every 1.00 kg of CO2(g) produced on burning the following auto fuels.
a. octane, cH° = –5074.1 kJ/mol. (14.4 MJ)
b. ethanol,cH° = –1234.8 kJ/mol. (14.0 MJ)
Use the following information to answer the next question The data below were obtained using a bomb calorimeter for the efficient
complete combustion of several samples of a liquid organic compound.
Mass of Liquid Compound (g) Energy Released (kJ)
3.35 66.0
3.50 69.1
3.25 63.6
3.30 64.5
11. Preliminary investigation suggests that the liquid is either ethanol or methanol. Which alcohol was burned?
Your answer should also include
the average energy produced by the liquid compound in kJ/g
Use the following information to answer the next question
In 2005 Volvo launched a truck powered by a DME, dimethyl ether or (CH3)2O(g), engine. The truck has a conventional diesel engine with a fuel system modified for DME. This unbalanced equation represents the DME.combustion.
__ (CH3)2O(g) + __ O2(g) __ CO2(g) + __ H2O(l)
The DME can be produced from biomass (wood, garbage, and waste vegetable matter) or from carbon dioxide extracted directly from the air.
H
f
(DME) = –184.1 kJ/mol H (Diesel) = 42.7 MJ/kg
12. Compare DME to diesel for powering trucks. Your answer should include
an enthalpy change the balanced combustion of DME
the energy to mass ratio (MJ/kg) of DME
Heater Efficiency
Many factors impact the quantity of energy absorbed by water being heated by a
chemical reaction that revolve around the design of the heater apparatus and the sources of heat loss inherent in all heating devices. Simply put, heater efficiency is really the percentage yield of a combustion or other heating process as follows:
100%
Heat absorbed by water
Heater Efficiency
Heat released by fuel combustion
Conveniently, this can be rewritten as
( )
% heater 100%
c fuel
m c t
Efficiency
n H
Here (m c t)heater is the energy absorbed by the water within the heating device – the
measured heat release. In the ideal world this would be all the energy released by the combustion reaction …clearly an unrealistic scenario. Obviously, heat is lost to the immediate surroundings of the heater, the burner assembly, the containers, and in the hot flue gases that we vent to the outside. n cH fuel is the predicted heat released by the burning fuel
Example : Calculate the efficiency of the hot water heater that warms 200 kg of water from 5 to 45°C, burning up 1.00 kg of methane in the process. (Burning one mole of methane to give gaseous products releases 802.5 kJ of heat energy.)
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) cH° = –802.5 kJ
4 3
4
( 200 4.19 / 40 )
% 100%
802.5 1.00 10
16.05
67 % .
o o
kg kJ kg C C
Efficiency
mol CH kJ
g
g mol CH
efficient
A typical, gas fueled, basement hot water heater is shown opposite
Gas hot water heaters, depending on design, have efficiencies in the range 60 – 90 %.
Oil fired hot water heaters are ca 50 – 60 5 efficient. Electric hot water heaters are 90 – 95% efficient. (Heating element within the water)
From an environmental perspective electrical heating is not the best. Coal or natural gas was burned to vaporize water, to drive turbines and dynamos to generate the electricity! Such
processes are ca 60% efficient.
1. In a calorimetric experiment, a student found that the theoretical value for the molar enthalpy of combustion for ethanol was higher than the value that she obtained in the laboratory. What is the most likely explanation for this discrepancy?
3. The calorimetric apparatus below was used by a student to determine experimentally the molar enthalpy of the combustion of candle wax, C25H52(s).
a. List all the measurements that must be taken in order to determine the molar enthalpy of combustion of the candle wax.
b. Write a mathematical equation that uses the data collected and that will allow you to determine the molar heat of combustion. Label each of the mathematical variables used in the equation.
c. Suggest two improvements to the experimental design.
When particles react the way in which they collide (collision geometry) and how much energy they have ( activation energy) are very important
Particles must hit in such a way that a reaction is favorable and with sufficient force – which is different for every chemical
reaction. Think of Lego. You must have the correct orientation and force to build something. Try putting some Lego in a bag and shaking it. Did you just make a castle? Now give it to your brother he is the catalyst he can orient the Lego and provide the force needed to connect them. Voila! A castle!
Back to Lego ---if you start with Lego pieces and need to rearrange them before you build a new “product” you have to add some energy to break apart the first blocks and start forming the products - --the is the activation energy required to complete he process. Activation energy is the minimum amount of energy needed to convert reactants to products.
How can we speed up the process?
Heat it up ---by heating up the particles we are increasing their kinetic energy --- giving them more velocity –this increases the probablility that they will collide with the correct orientation per unit time as well as have enough energy to produce the activated complex. We can see this effect by looking at the Maxwell-Boltzman distribution graph
As the reactants collide, chemical bonds break and form.
Breaking bonds is endothermic and Bond forming is exothermic
This works for exothermic and endothermic reactions. Catalysts can be inorganic, like Pt(s) or organic, like catalase, amylase, protease, which are biological enzymes. They do not get used up in the process so they can be used over and over again.
The overall enthalpy change for catalyzed and uncatalyzed reactions does not change. The only difference is the input energy, or energy required for the reaction is lowered in a catalyzed reaction…show as a lower transition state or activated complex.
Activation energy can also be looked at in the forward and reverse direction of both exothermic and endothermic reactions
Drawing a Potential Energy Diagram
Sample Problem:
Carbon monoxide, CO(g), reacts with nitrogen dioxide, NO2(g). Carbon dioxide, CO2(g), and nitrogen monoxide, NO(g), are formed. Draw a potential energy diagram to illustrate the progress of the reaction. (You do not need to draw your diagram to scale). Label the axes, the transition state, and the activated complex. Indicate the activation energy of the forward reaction, Ea(fwd) = 134 kJ, as well as ΔrH = –226 kJ. Calculate the activation energy of the reverse reaction, Ea(rev), and show it on the graph. Because the reaction is going in the reverse direction, ΔrH would be positive.
Solution
The activation energy of the reverse reaction may be determined using the formula described on page 407, or it can be determined from the potential energy diagram. The reaction is exothermic; therefore, the reverse reaction is endothermic.
Ea(rev) = Ea(fwd) + ΔrH = 134 kJ + 226 kJ
= 360 kJ
The activation energy of the reverse reaction is 360 kJ.
Look carefully at the potential energy diagram. Check that you have labelled it
completely. Since the forward reaction is exothermic, the reactants should be shown at a higher energy level than the products, and they are. The value of Ea(rev) is reasonable.
Draw and label potential energy diagrams to answer the following questions:
1. Assume the following reaction has a one-step mechanism: A + B C
Draw a potential energy diagram for the reaction that correctly incorporates each of the following labels:
E = 150 kJ/mol, Ea(fwd) = 450 kJ Potential energy, Reaction progress Transition state, A + B
C
(a) Is the reaction endothermic or exothermic?
(b) Determine Ea(rev) for the reaction. Add this value to your diagram.
(c) How would your diagram have been different if the reaction had a two-step mechanism? Explain your answer.
2. Assume the following reaction has a one-step mechanism: D + E F + G
Draw a potential energy diagram for the reaction that correctly incorporates each of the following labels:
∆E = –56 kJ/mol, Ea(rev) = 120 kJ Potential energy, Reaction progress Transition state, D + E
F + G
(a) Is the reaction endothermic or exothermic?
(b) Determine Ea(fwd) for the reaction. Add this value to your diagram.
Solar Energy and Fossil Fuels.
The Carbon cycle is shown below in simplified form. The arrows depict the flow of
carbon dioxide to and from the air and the oceans. Some plant and animal material is diverted to form fossil fuels, coal and petroleum/natural gas, respectively. Increased levels of carbon dioxide in the atmosphere, it is a greenhouse gas, are responsible for our current Anthropogenic Greenhouse Effect, a central part of our current concerns usually summed up as climate change. Since the vast majority of the energy used to fuel our economies in the west comes from burning fossil fuels in ever larger amounts the production of CO2(g) is of major concern.
Photosynthesizers, algae in the top 1-2 m of our oceans, certain bacteria, and plants on land, absorb solar energy, carbon dioxide, and water and convert it into carbohydrate. The net reaction for this process is often represented as:
Solar energy + 6 CO2(g) + 6 H2O(g) C6H12O6(aq) + 6 O2(g)
Photosynthesis, like cellular respiration also a biological redox reaction, is crucial for life on Earth. As well as maintaining the normal level of oxygen in the atmosphere, nearly all life either depends on it directly as a source of energy (plants and herbivores) or
indirectly as the ultimate source of the energy in their food (omnivores and carnivores). Photosynthesis provides the source materials that, with burial and heating under pressure over geological ages, produced fossil fuels.
Burning fossil fuels is essentially exploiting solar energy stored, in some cases, up to 300 million years ago. During the Carboniferous Period, from about 359 to 299 Mya, many beds of coal were laid down all over the world. Alberta had many dense and swampy forests during this period, which gave rise to large deposits of peat. Since that time, with the aid of burial, heat, and pressure, the peat transformed into rich coal stores. The name "Carboniferous" refers to this coal.
Cellular respiration, in the mitochondria of all organisms, convert carbohydrate and oxygen into carbon dioxide and water with the release of metabolic energy – some of which goes toward temperature maintainance, and the rest to producing ATP. The net reaction for this biological redox process is often represented as:
C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + metabolic energy
To be more precise the state of the reactants and products in the cell respiration net reaction are more likely aqueous within the confines of a cell. Nevertheless the enthalpy change of this process, ca -2800 kJ, calculated using formation enthalpies for the
components in the states above, is considered little different from the true value. Thus, cellular respiration is a means of converting the chemical energy resulting from
photosynthesis into more useful forms. Combustion of glucose, or other foods, in a bomb calorimeter is considered a good measure of the amount of energy obtained in the foods we eat. The following diagram illustrates how food is ultimately converted into heat energy.
Respiration
Foodstuffs + O2(g) ATP + CO2(g) + H2O(l) + Heat
Cell Work
Heat
In simple terms the net reaction for cellular respiration is the reverse of the net reaction for photosynthesis.
Photosynthesis
