Thermochemical Equations
A chemical equation may look like the following: CH4 + 2O2 → CO2 + 2H2O
A THERMOCHEMICAL equation looks like the following:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔH = –890kJ
What is different?
¾ States are ____________ specified: (s) solid, (l) liquid, (g) gas, (aq) aqueous.
¾ ΔH values (usually kJ mol-1) are signed (+/–).
¾ ΔH is a _______________________ term.
This thermochemical equation tells us that when 1 mole of methane is burned
completely in air to give CO2 gas and liquid water, 890 kJ of energy is given off.
Convention: Measurements usually made at 25°C (298K).
A typical ENDOTHERMIC reaction: CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ
i.e., we must supply 178 kJ of energy to convert 1 mole CaCO3(s)
to CaO(s) and CO2(g).
Some Simple Thermochemical Rules & Laws
Thermochemical equations normally tell you about energy changes
associated with _______________ of something.
e.g., We saw above that the ΔH for combustion of 1 mole CH4
at 298 K was –890 kJ.
When a thermochemical equation is multiplied by any factor,
the value of ΔH is ____________________________.
Question: How much energy is released if we burn 1.00 L CH4(g) at
1.01×105 Pa and 25°C?
The first thing we should ask ourselves is "How many moles of
CH4 is present in 1.00 L at 1.01×105 Pa & 25°C?"
Ideal Gas Law:
Note: If P in Pa and V is m3, then use R = 8.314 Pa·m3·K–1 mol–1
If P in kPa and V is L, then us R = 8.314 kPa·L·K–1 mol–1
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# moles of CH4(g) at 1.01×105 Pa and 25°C:
Since PV = nRT, therefore n = PV/RT
n=
Based on the thermochemical equation given earlier, 1 mole CH4
evolves __________ kJ, therefore 4.08×10–2 moles of CH4 evolves
4.08×10–2 mol × __________ kJ/mol = 36.3 kJ
If we REVERSE the direction of a reaction, we _________________ of ΔH
e.g. CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ
CaO(s) + CO2(g) → CaCO3(s), ΔH = _______________ kJ
Thermochemistry (and Thermodynamics) is only interested in energy
differences between initial and final states. It is not concerned with HOW
(i.e., "by what mechanism") a reaction takes place. All this means is that we
can add (or subtract) thermochemical equations to produce other equations.
Hess's Law: Thermochemical equations can be added (or subtracted) to
yield other thermochemical equations. (more on this later)
THERMOCHEMISTRY OF CHANGE
1. Changes of Physical State
e.g., solid → liquid → gas
solid → gas
OR
A. Enthalpy of Fusion, ΔHfus: the enthalpy change when we fuse (i.e., melt) 1
mole of something.
i.e., 1 mole solid →
e.g.,
ΔH°fus is always ______________________; energy had to be added to
weaken forces (_______________) and cause the solid to flow.
Note: T remains constant while process occurring, e.g., 0°C for H2O(s).
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The "°" symbol refers to “STANDARD CONDITONS”:
• pure reactants at 1 atm of pressure (actually 1 bar which equals 105 Pa)
•
pure products at _____________________________.
•
solids & liquids are in their most stable form, known as "standard state"
•
solutes are ________________________.
•
temperature must be specified.
Note: The most common temperature used is 25°C BUT it is not included
within the definition of "standard" conditions.
For the reverse process:
H2O(l) → H2O(s); ΔH° =
0°C
0°C
What is the name for this process? _________________________________
i.e., In general, ΔHReverse process = –ΔHForward process
B. Enthalpy of Vaporization, ΔHvap enthalpy change when we vapourise 1
mole of liquid to gas. (ΔH°vap is always _________________)
e.g.,
H2O(l) → H2O(g); ΔHvap =
100°C
100°C
What is the name for the reverse process?
C. `Enthalpy of Sublimation, ΔHsub:
– enthalpy change when we sublime 1 mole solid → 1 mole _____________
(ΔH°sub is always ______________________)
e.g.,
Note: ΔH°sub = ΔHfus + H°vap ...can do this since enthalpy is a state function.
2. Enthalpy Changes for Chemical Reactions
A. Enthalpy of Combustion, ΔHcomb:
– enthalpy change when we burn ____________________ completely in O2.
e.g., C(s) + O2(g) → CO2(g); ΔHcomb = – 393.5 kJ
ΔHcomb is always ___________________, because energy is always
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_____________________ (exothermic).
B. Enthalpy of atomization:
– enthalpy change associated with the formation of _____________________
from its elements.
e.g., ½H2(g) → H(g); ΔH° = +218 kJ
Note: Enthalpy of reaction or "Heat" of reaction is a general description
used to refer to any reaction.
HESS'S LAW
(Refer to page 241 in text.)
Hess's Law – ΔH for an overall process is the sum of the ΔH's for
the individual steps of the process.
Thermochemistry (and thermodynamics) is only interested in the energy
differences between initial and final states. It doesn't matter how we got
there. Hence we can add (or subtract) thermochemical equations to produce
other thermochemical equations.
Example:
Suppose we wanted to know ΔH°Rxn for Sn(s) + 2Cl2(g) → SnCl4(l), but you could
not find the value in any set of tables. But, you did find the following:
(a) Sn(s) + Cl2(g) → SnCl2(s); ΔH°Rxn = –350 kJ
(b) SnCl2(s) + Cl2(g) → SnCl4(l); ΔH°Rxn = –195 kJ
Can you manipulate these equations to arrive at the one we want?
Sn(s) + 2Cl2(g) → SnCl4(l),
Example: Given the following data:
N2(g) + 3 O2(g) → N2O3(g)
2
N2(g) + O2(g) → 2NO(g)
½N2(g) + O2(g) → NO2(g)
what is ΔH for the rxn:
A) –207.1 kJ
B) –39.7 kJ
C)
24.3 kJ
ΔH°Rxn = __________ kJ
ΔH = 83.7 kJ
ΔH = 180.4 kJ
ΔH = 33.2 kJ
N2O3(g) → NO(g) + NO2(g)
D) 39.7 kJ
E) 207.1 kJ
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Problem:
Given:
(a) ½ N2(g) + ½ O2(g) → NO(g); ΔH°Rxn = +90.4 kJ
(b) ½ N2(g) + O2(g) → NO2(g); ΔH°Rxn = +33.2 kJ
Calculate ΔHRxn for: 2NO(g) + O2(g) → 2NO2(g)
Ans: ΔHRxn= – 114.4 kJ
To use Hess's Law effectively we obviously need to have access to tables of ΔH
values. Rather than write out full thermochemical equation for every ΔH, we devise
a method (outlined below) which will allow you to deduce the
appropriate thermochemical equation.
STANDARD ENTHALPY OF FORMATION (ΔH°f)
The Standard Molar Enthalpy of Formation (or the standard heat of
formation) is the enthalpy change when we form ONE mole of a substance
from its ELEMENTS in their STANDARD STATES.
Aside: Previous, we looked at the combustion reaction:
C(s) + O2(g) → CO2(g); ΔHcomb = – 393.5 kJ
This reaction also leads to the formation of CO2(g) in its standard state.
Hence, the heat of this reaction (–393.5 kJ/mol), which we noted as the
heat of combustion, is also the heat of formation of CO2(g), i.e.,
ΔH°f [CO2(g)] = –393.5 kJ/mol
These ΔH°f values can be obtained for many, many compounds (see Appendix C:
Thermodynamic Quantities for Substances and Ions at 25°C). ΔHf° can have either
negative or positive values, e.g.,
Different forms of bromine ΔH°f (kJ/mol)
Br2(l)
Br2(g)
Br(g)
0
30.91
111.88
Note: The energy difference between Br2(g) and Br(g) corresponds
to the energy to break the Br-Br bond.
Allotropes of Carbon ΔH°f (kJ/mol)
C(s, graphite)
C(s, diamond)
0
+1.897
You should be able to write an equation that defines a heat of formation.
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Example: The heat of formation (ΔH°f) for liquid ethanol, C2H5OH(l) is –
278 kJ/mol, write the thermochemical equation.
→
Step 1:
(form 1 mole of the compound)
→
Step 2:
(from its elements)
→
Step 3:
ΔHf° =
kJ (balance eq’n)
Note: For a pure element in its most stable form, at a chosen temperature
(usually 25°C), ΔH°f = 0 J at 1 atm.
WHY?
Because: O2(g) → O2(g); ΔHf° = 0
From a list of ΔH°f we can:
¾ Calculate heats of other reactions
¾ Calculate ΔH°f’s that are difficult to measure
In some cases, ΔH°f cannot be defined directly by writing an equation from
the elements. Therefore, must combine other equation to get the desired
products (see below).
Question: The standard enthalpy of formation of sulfuric acid, H2SO4(l), refers to
the ΔH° of which of the following reactions?
A) H2SO4(l) → H2(g) + S(s) + 2O2(g)
B) H2(g) + S(g) + 2O2(g) → H2SO4(l)
C) H2SO4(l) → 2H(g) + S(s) + 4O(g)
D) H2(g) + S(s) + 2O2(g) → H2SO4(l)
E) 2H(g) + S(g) + 4O(g) → H2SO4(l)
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Problem:
Given ΔHf° values for H2O(l) = –286 kJ/mol
H2O(g) = –242 kJ/mol
Calculate ΔH°vap for H2O(l)
We want H2O(l) → H2O(g),
ΔH°vap = ?
We are given:
(a) H2(g) + ½O2(g) → H2O(l),
(b) H2(g) + ½O2(g) → H2O(g),
Apply Hess's Law:
ΔH = –286 kJ
ΔH = –242 kJ
H2O(l) → H2O(g),
ΔH°vap = __________ kJ
∴ ΔH°vap for H2O(l) is ____________ kJ/mol
i.e., 44.0 kJ must be SUPPLIED to evaporate 18.0 g H2O(l) → 18.0 g H2O(g)
What we have just seen is an example of a general rule IF we have ΔHf° data:
ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)
ANOTHER APPROACH:
2CH3CHO(l) → C2H5OH(l) + CH3COOH(l)
ethanal
ethanol
ethanoic acid
(acetaldehyde)
(acetic acid)
Envisage a hypothetical process:
Step 1: Reactants → Constituent Elements
Step 2: Constituent Elements → Products
For Step 1: ΔH° = sum of –ΔHf° for all reactants
For Step 2: ΔH° = sum of ΔHf° for all products
Combining the two: ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)
Returning to the original equation:
2CH3CHO(l) → C2H5OH(l) + CH3COOH(l)
Given
ΔHf° values –192.3
–277.69
–484.5
kJ/mol
ΔH°rxn = [–277.69 – 484.5] – [–192.3 × 2]
ΔH°rxn = –377.6 kJ
Note: Watch to make sure you include the coefficients within the balanced equation
(e.g., 2CH3CHO) to arrive at the correct amount of energy.
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Remember: Any equation showing the standard molar enthalpy of formation of a
compound must show the formation of 1 mole from the elements in their
standard states, e.g., ΔH°f (NH3) is –46.11 2 kJ/mol
The thermochemical eq’n is then: ½N2(g) + 3/2H2(g) → NH3(g); ΔH°f = –46.11 kJ
When using ΔH°f values – you have two ways to choose from:
1) long way (Hess' law)
2) short way (using ΔH°f directly)
Question: Calculate the heat required to decompose limestone:
CaCO3(s) → CaO(s) + CO2(g)
You are given:
ΔH°f [CaCO3(s)] = –1206.9 kJ/mol
ΔH°f [CaO(s)]
= –635.1 kJ/mol
ΔH°f [CO2(g)]
= –393.5 kJ/mol
1) The Long Way (Hess's Law):
First write out the three thermochemical equations for the three given
heats of formation:
a)
b)
c)
Then see how to manipulate them to arrive at needed thermochemical equation:
For the given reaction, CaCO3(s) appears on the LHS, therefore
equation a) must ________________________________
CaO(s) and CO2(g) must appear on the RHS, so equation b) must
__________________________________
CaO(s) and CO2(g) must appear on the RHS, so equation c) must
_________________________________.
So, rewrite the equations and add them up:
__________________________________________________________
CaCO3(s) → CaO(s) + CO2(g);
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ΔH°Rxn = +178.3 kJ
2) The Short Way
The mathematics of the above problem really amounts to the same as,
ΔH°rxn = ΣnΔHf°(products) – ΣnΔHf°(reactants)
In this case,
ΔH°rxn =
ΔH°rxn =
ΔH°rxn = +178.3 kJ
So, if you have ALL the ΔHf°data you can simply do it the short way (a shortcut around
writing down all the thermochemical equations). Note, however, Hess's Law works ALL
the time.
Problem 1:
Calculate ΔH°Rxn for the reaction CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Ans: ΔH°Rxn = –891 kJ/mol
Problem 2:
Given
(a) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l); ΔH = –1169 kJ
(b) 4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l); ΔH = –1530 kJ
What is ΔH°f for NO(g)?
Ans: ΔH°f = +90.3 kJ/mol
Problem 3:
Given ΔHf°[C2H6(g)] = –85 kJ/mol
ΔHf°[H2O(l)] = –286 kJ/mol
ΔHf°[CO2 (g)] = –394 kJ/mol
Calculate how much energy is released when 1.00 g C2H6 (g) is burned in air at 25°C and
1.01 × 105 Pa.
Ans: 1.00 g C2H6 evolves 51.9 kJ of energy upon combustion.
Example from Nature:
The Bombardier Beetle defends itself by "shooting" a hot solution of benzoquinone at its
enemies.
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Reaction: C6H4(OH)2(aq) + H2O2(aq) → C6H4O2(aq) + 2H2O(l)
(a) Calculate ΔH°Rxn for this reaction given:
(i) C6H4(OH)2(aq) → C6H4O2(aq) + H2(g); ΔH°Rxn = +177 kJ
(ii) ½O2(g) + H2O(l) → H2O2(aq); ΔH°Rxn = + 94 kJ
(iii) H2(g) + ½O2(g) → H2O(l); ΔH°f = –286 kJ
Ans: –203 kJ
(b) Calculate the rise in temperature of 4.0 mg H2O produced by oxidation of 8.0 × 10–6
mol C6H4(OH)2.
Ans: ΔT = 97C° ---> hot stuff!
STANDARD ENTHALPIES OF FORMATION
OF IONS IN AQUEOUS SOLUTION
To deduce the ΔHf° values for all ions in aqueous solution, we needed a method
to separate the contributions of the ions.
SOLUTION: Define ΔHf°(H+(aq)) = 0
****************************************************************
ASIDE: Consider the reaction: HCl(g) → H+(aq) + Cl–(aq)
ΔH°Rxn = {ΔHf°[H+(aq)] + ΔHf°[Cl–(aq)]} – {ΔHf°[HCl(g)]}
(measured)
(obtain)
(known)
•
ΔH°Rxn for this reaction is measured as –74.89 kJ/mol
•
ΔHf°(HCl(g)) is known as – 92.31 kJ/mol
We can then write:
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ΔH°Rxn = {ΔHf°[H+(aq)] + ΔHf°[Cl–(aq)]} – {ΔHf°[HCl(g)]}
–74.89 = {ΔHf°[H+(aq)] + ΔHf°[Cl–(aq)]} – (– 92.31)
{ΔHf°[H+(aq)] + ΔHf°[Cl–(aq)]} = –167.20 kJ/mol
Although we can obtain the SUM this way, the separate contributions of the
two ions cannot be obtained by any experiment yet devised.
What to do?
SOLUTION: Define ΔHf°(H+(aq)) = 0
Hence for the reaction: ΔHf°[Cl–(aq)]} =
Now that we have ΔHf°[Cl–(aq)]} we can get ΔHf°[Na+(aq)]} from an experiment
using NaCl(s) (i.e., NaCl(s) → Na+(aq) + Cl–(aq))
From similar experiments we can determine the values for other ions
(e.g., NaBr, KBr, etc.)
Therefore, we can deduce the ΔHf° values for all other ions in aqueous solution.
*******************************************************************
Questions
1. Which is/are endothermic process(es)?
1. Combustion
2. Fusion
A. 1 only
B. 2 & 3 only
3. Vapourisation
C. 2, 3 & 4 only
4. Sublimation
D. 2, 3, 4 & 5 only
5. Atomisation
E. All of the above!
2. Which of the following has (have) a non-zero enthalpy of formation value at 25°C and
1.00 atm?
b. O2(g)
c. O3(g)
a. O(g)
A)
a only
D)
a and c only
B)
b only
E)
a, b, and c
C)
c only
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