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Wize High School Grade 12 Chemistry Textbook > Energy Changes

Ways to Calculate Enthalpy of Reactions

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Hess' Law

We will be looking at 2 ways to calculate ΔHorxn:

1) Hess' Law of Formation Method

2) The Heats of Formation Method (ΔHof)


  • If a reaction is carried out in a series of steps, ΔH for the overall reaction can be found from the sum of the enthalpy changes of the individual steps.
  • This is because enthalpy (H) is a state function and when it changes, it does not depend on the pathway taken
ΔHrxn = ΔH1 + ΔH2 + ΔH3 + …


Consider the following thermochemical equations:
C (s) + O2 (g) → CO2 (g) ΔH = -393.5 kJ
2 C (s) + O2 (g) → 2 CO (g) ΔH = -221.1 kJ
The ΔH for the reaction: CO (g) + ½ O2 (g) → CO2 (g) is what?

Solution:

1◯  C(s)+O2(g)→CO2(g)=−393.5 KJ2◯2C(s)+O2(g)→2CO(g)=−221.1 KJ\text{\textcircled 1} \space \space C (s) + O_2 (g) → CO_2 (g) = -393.5 \space KJ\\\text{\textcircled 2} 2 C (s) + O_2 (g) → 2 CO (g) = -221.1 \space KJ

Equation 1 can stay the same (we see that it has CO2 on the right side of the equation and in the overall equation we want CO2 to be on the right side. The coefficient of CO2(g) does not need to be changed either).

For equation 2, we see that it has 2CO on the right. The overall equation has 1 CO on the left. This means that we would need to flip equation 2 (multiple deltaH by -1 and multiply equation 2 by 1/2. In total for equation 2 we will multiply it by -1/2.

1) C(s) + O2(g) --> CO2(g) deltaH= -393.5kJ
2) CO(g) --> C(s) +1/2O2(g) deltaH= +110.6kJ

Now for the overall reaction, add 1) and 2):
CO(g) + 1/2O2(g) --> CO2(g) deltaH= -282.9kJ


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Hess' Law


Wize Concept
The 2 things you need to remember for this method are:

1) Whenever we want to look at a reaction in reverse, we must multiply ΔH by -1
2) Whenever we want to multiply a reaction step by a coefficient, we must multiply ΔH of that step by that same coefficient!


What is the ΔHrxn for the following reaction?


2C2H6(g) + 7O2(g) --> 4CO2(g) + 6H2O(g)


1) 2C + 3H2 --> C2H6 ΔH=-84.68kJ

2) C + O2 --> CO2 ΔH=-394kJ

3) H2 + 1/2O2 --> H2O ΔH=-286kJ


The following reaction enthalpies are provided:
H2(g)+12O2(g)→H2O(l),ΔH1=−285.8kJ/mol(1)H_{2(g)}+\frac{1}{2}O_{2(g)}\to H_2O_{(l)},\Delta H_1=-285.8kJ/mol\hspace{70pt} (1)
C(s)+O2(g)→CO2(g),ΔH2=−293.5kJ/mol(2)C_{(s)}+O_{2(g)}\to CO_{2(g)},\Delta H_2=-293.5kJ/mol \hspace{80pt} (2)
CH4(g)+2O2(g)→CO2(g)+2H2O(l),ΔH3=−890.4kJ/mol(3)CH_{4(g)}+2O_{2(g)}\to CO_{2(g)}+2H_2O_{(l)},\Delta H_3=-890.4kJ/mol \hspace{17pt} (3)

Use this data to find the ΔHrxn for the following reaction:
C(s)+2H2(g)→CH4(g)C_{(s)}+2H_{2(g)}\to CH_{4(g)}
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Heats of Formation

We will be looking at 2 ways to calculate ΔHorxn:

1) Hess' Law of Formation Method

2) The Heats of Formation Method (ΔHof)


  • Heat of formation is the amount of heat that is required to form 1 mole of a compound from its constituent elements in their natural/standard state (the state they are in under standard conditions: P=
    1
    atm, T=
    25
    oC,
    1
    M concentration, pH
    7
    .
  • The ΔHof=0 for an element in standard state

Wize Concept
**The following elements in their standard state should be memorized and the phase they are in is important too!
Diatomic molecules (BrINClHOF)
  • For example, Cl2(g), H2(g), O2(g).
C(s) as graphite is in standard state.
I2(s)
Br2(l)
Hg(l)

Watch Out!
If you are shown O(g) on your exam, this is NOT oxygen in its standard state. Remember oxygen in its standard state has to be gaseous AND diatomic! So: O2(g) would be in standard state.


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Example: The formation reaction for C4H10O(l) would be:

4C(s)+5H2(g)+12O2(g)→C4H10O(l)4C_{(s)}+5H_{2(g)}+\frac{1}{2}O_{2(g)}\to C_4H_{10}O_{(l)}
  • In fact, if you know the enthalpy of formation for each reactant and product in any chemical equation, you can find the enthalpy change for that reaction with the following formula:

ΔHorxn=[(∑nΔHofproducts)−(∑nΔHofreactan⁡ts)]\Delta H^orxn=\left[\left(\sum_{ }^{ }n\Delta H^of_{products}\right)-\left(\sum_{ }^{ }n\Delta H^of_{reac\tan ts}\right)\right]

  • This equation says to add up the enthalpies of formation for each product, multiplied by the stoichiometric coefficient of that product, and subtract it from the sum of the enthalpies of formations of the reactants multiplied by their coefficients.
  • In other words, Δ= Final - Initial

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Example:
2NO(g) + O2(g) eqm 2NO2(g) Calculate ΔHorxn.
ΔHof(NO(g))=90.75kJ/mol
ΔHof(NO2(g))=33.18kJ/mol
ΔHof(O2(g))=
0


ΔHorxn=[(∑nΔHofproducts)−(∑nΔHofreactan⁡ts)]\Delta H^orxn=\left[\left(\sum_{ }^{ }n\Delta H^of_{products}\right)-\left(\sum_{ }^{ }n\Delta H^of_{reac\tan ts}\right)\right]
ΔHorxn=[(2x33)-(2x90 + 0)]
ΔHorxn=[(66-180)]
ΔHorxn= -114 kJ

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Write a reaction for the heat of formation for dimethylaminopyridine, (CH3)2NC5H4N.


7 C(s)+5 H2(g)+N2(g)→(CH3)2NC5H4N(s)7\ C_{(s)}+5\ H_{2(g)}+N_2(g)\rightarrow(CH_3)_2NC_5H_4N_{(s)}
Recall: C in its standard state is in the form of C(s) or C(graphite)
Note: In a formation reaction only want to form one mole of the substance!
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Find ΔHrxn for the following reaction at 25 °C:

C6H10O3(l)+7O2(g)→6CO2(g)+5H2O(l)C_6H_{10}O_{3\left(l\right)}+7O_{2\left(g\right)}\to6CO_{2\left(g\right)}+5H_2O_{\left(l\right)}
Given the following data:
ΔHrxn=(∑nproductΔHof)−(∑nreactantΔHof)\Delta H_{rxn}=\Bigg(\sum n_{product}\Delta H^of\Bigg)-\Bigg(\sum n_{reactant}\Delta H^of\Bigg)
=(6(−393.5kJ/mol)+5(−285.8kJ/mol))−(−640.4kJ/mol)=\bigg(6(-393.5kJ/mol)+5(-285.8kJ/mol)\bigg)-(-640.4kJ/mol)
Note that O2(g) is already in its standard elemental state and so it doesn’t need to be included in the calculation (ΔHof (O2(g))=0) .

Therefore, ΔHrxn is -3150 kJ/mol. Since this is a combustion reaction, we expect a negative ΔH value, in agreement with calculations.
Calculate the ΔH°f value of benzene (C6H6) from the following information:

2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(l) ΔH° = −6535 kJ/mol
CO2(g) ΔH°f = −393.5 kJ/mol
H2O(g) ΔH°f = −241.8 kJ/mol
H2O(l) ΔH°f = −285.8 kJ/mol


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2H2(g) + O2(g) --> 2H2O(l) ΔH=-572kJ/mol


1. How much heat is produced when 72g of water gas is produced?

Currently we see that when 2 mol of H2O(l) are produced we have 572kJ/mol of heat produced.
We need to find out how many moles of water we have when we have 72g of water:

n=? m=72g, M=18g/mol
n=mMn=\frac{m}{M}

n=72g18gmoln=\frac{72g}{\frac{18g}{mol}}
n=4 moles
Now that we have 4 moles of H2O(l) produced (double the amount shown in the equation), that means that the heat produced would also be doubled.)
Therefore, 1040kJ of heat would be released (q=-1040kJ) when 4 moles of water is produced.

2. Is the reaction endothermic or exothermic?

Since ΔH is negative, it means heat is being released and the reaction is therefore exothermic (heat is EXiting!)

3. We are told that O2(g) is the limiting reagent of this reaction. How many moles of O2(g) are used if the reaction releases 286kJ of heat?

  • 286/572=1/2
  • Since we now have half the amount of heat released, it means that we would start with half the amount of moles of O2 that we have in the reaction
  • The reaction has 1 mol of O2 leading to 572kJ of heat being released
  • So 1/2 a mole of O2 would lead to 286kJ of heat being released
Given this reaction, answer the following questions:

C3H8(g) + 5O2(g) --> 3CO2(g) + 4H2O(l)


What is the ΔHo for the reaction? (Ans in kJ and round to the nearest whole number)
Extra Practice
Enthalpy of Formation
What is the enthalpy of the following reaction when 0.35 g of Mg2+ is produced? Use the following information:
ΔH°f[Mg(OH)2(s)] = −924.5 kJ
ΔH°f[NH4+(aq)] = −132.5 kJ
Gibbs Free Energy
The reaction of
H2(g)+CO2(g)↔H2O(g)+CO(g) at 2000K KP=4.40H_{2(g)}+CO_{2(g)}\leftrightarrow H_2O_{(g)}+CO_{(g)}\ at\ 2000K\ K_P=4.40
Calculate ΔG for this reaction
Thermochemistry: Enthalpy of reactions
The following reaction enthalpies are provided:
H2(g)+12O2(g)→H2O(l),ΔH1=−285.8kJ/mol(1)H_{2(g)}+\frac{1}{2}O_{2(g)}\to H_2O_{(l)},\Delta H_1=-285.8kJ/mol\hspace{70pt} (1)
C(s)+O2(g)→CO2(g),ΔH2=−293.5kJ/mol(2)C_{(s)}+O_{2(g)}\to CO_{2(g)},\Delta H_2=-293.5kJ/mol \hspace{80pt} (2)
Calorimetry: Temperature change
1. A 56.0 g block of iron heated to 90 °C is placed in a 4.8 kg water bath at 20 °C. What will be the final temperature of the iron? heat capacity of iron c = 0.450 J /gK, heat capacity of water is 4.185 J/gK
Enthalpy of Reactions
Consider the following reaction:
N2H4(I)+O2→N2(g)+2 H2O(I)ΔH=−622.2 kJN_2H_4(I)+O_2\rightarrow N_2(g)+2\ H_2O(I)\quad \Delta H=-622.2\ kJ
Given the following data, calculate the heat of reaction for the same reaction where water is a gaseous product instead of a liquid:
Calculate ΔH°\Delta H\degree for the reaction below, which requires catalytic acid to proceed
C6H12(l)+H2O(l)→C6H13OH(l)C_6H_{12}\left(l\right)+H_2O\left(l\right)\to C_6H_{13}OH\left(l\right)
Given:
The standard enthalpy of formation for NO2(g)NO_2\left(g\right) is 33.18 kJ/mol, and the standard enthalpy of partial combustion of nitrogen to NO(g) is 90.25 kJ/mol. Calculate the standard enthalpy of combustion of NO(g) to NO2(g).
Hess’s Law: Indirect Determination of ΔH
The standard enthalpy of formation of ethanol (C2H5OH){\small \left(\text{C}_2\text{H}_5\text{OH}\right)} is -277.7 kJ/mol. Write the chemical equation to which this value applies.
Solution:
Calculate the ΔH for the reaction below IF5 (g) → IF3 (g) + F2 (g)
IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ IF (g) + 2 F2 (g) → IF5 (g) ΔH = -745 kJ
Liquification of fuels is an important process that allows fuel to be transported more safely and more easily around the world. Oxidizing methane transforms it from a gas to a liquid, methanol, which is much easier to transport. Using the data provided calculate ΔH\Delta Hfor the oxidation of methane (CH4) shown below
CH4(g) + 1/2 O2(g)→CH3OH(l)ΔHfo(CH4(g))=−74.9kJ mol−1ΔHfo(CH3OH(l)=−238.6 kJ mol−1\begin{array}{c} CH_{4(g)}\ +\ 1/2\ O_{2(g)} \rightarrow CH_3OH_{(l)} \\ \\ \Delta H^o_f(CH_{4(g)})=-74.9 kJ\ mol^{-1} \\ \Delta H_f^o(CH_3OH_{(l)}=-238.6\ kJ\ mol^{-1} \end{array}
Practice: Total Bond Energy of IF$_7$
The total bond enthalpy for F2 is 156.9 kJ/mol and the total bond enthalpy of I2 is 152.5 kJ/mol. Using the thermochemical equation below, what is the total bond enthalpy of IF7?
I2(s)+7 F2(g)→2 IF7(s)  ΔH=2669.2 kJ.molI_{2(s)}+7\ F_{2(g)}\rightarrow 2\ IF_{7(s)}\ \ \Delta H=2669.2\ kJ.mol
Enthalpy of Reactions
Given the following at 25°C and 1.00 atm:
ReactionΔH°12N2(g)+O2(g)→NO2(g)33.2kJN2(g)+2O2(g)→N2O4(g)11.1kJ\def\arraystretch{1.5}\begin{array}{cc}\rm Reaction&\Delta H\degree\\\cfrac{1}{2} N_2 (g)+O_2 (g)→NO_2 (g)&33.2 kJ \\N_2 (g)+2O_2 (g)→N_2 O_4 (g)&11.1 kJ\end{array}
Enthalpy of Reactions
Estimate the heat of reaction at 298 K for the reaction shown, given the average bond energies below.
Br2(g)+3F2(g)→2BrF3(g)Br_2 (g)+3F_2 (g)→2BrF_3 (g)
BondBond EnergyBr−Br193kJ/molF−F155kJ/molBr−F249kJ/mol\def\arraystretch{1.5} \begin{array}{c} \rm Bond &&&& \rm Bond \ Energy \\ \hline Br-Br &&&& 193 kJ/mol\\ F-F &&&& 155 kJ/mol\\ Br-F &&&& 249 kJ/mol\\ \end{array}
Enthalpy: Bond Dissociation Energies
Consider the following balanced chemical reaction and the bond dissociation energies for the partial combustion of hydrogen sulfide:
What is the enthalpy change ΔrH for this reaction?
Thermochemistry: Calorimetry
Consider the following balanced chemical equation:
2CℓF3(g)+2O2(g)→Cℓ2O(g)+3F2O(g)∆rH°=???2 CℓF_3(g) + 2 O_2(g) → Cℓ_2O (g) + 3 F_2O (g) \qquad ∆_rH° = ???
Substance△fH° @ 25°C (kJ/mol)CℓF3(g)??Cℓ2O(g)80.3F2O(g)–21.7F2(g)0\def\arraystretch{1.5} \begin{array}{|c|c|} \hline \rm Substance & △_fH° \ @ \ 25°C \ (kJ/mol) \\ \hline CℓF_3 (g) & ?? \\ \hline Cℓ_2O (g) & 80.3 \\ \hline F_2O (g) & – 21.7 \\ \hline F_2 (g) & 0 \\ \hline \end{array}
Thermochemistry: Calorimetry
Ethene gas (C2H4) reacts with fluorine gas according to the following balanced equation:
C2H4(g)+6F2(g)→2CF4(g)+4HF(g)ΔrH°=−3012 kJ/mol⋅rxn @ 25°CC_2H_4 (g) + 6 F_2 (g) → 2 CF_4 (g) + 4 HF (g) \qquadΔ_rH° = −3012\ kJ/mol·rxn\ @ \ 25°C
Substance△fH° @ 25°C(kJ/mol)C2H4(g)??F2(g)0CF4(g)–933HF(g)–273.3\def\arraystretch{1.5} \begin{array}{|c|c|} \hline \rm{Substance} & △_fH° \ @ \ 25°C(kJ/mol) \\ \hline C_2H_4 (g) & ?? \\ \hline F_2 (g) & 0 \\ \hline CF_4 (g) & – 933 \\ \hline HF (g) & – 273.3 \\ \hline \end{array}
Thermodynamics: Enthalpy of reaction
Find ΔHo for the following reaction if 12.0 g of N2H4 is consumed in excess N204.
2N2H4(l)+N2O4(g)→3N2(g)+4H2O(l)2N_2H_{4(l)}+N_2O_{4(g)}\to 3N_{2(g)}+4H_2O_{(l)}
The following enthalpies of formation are provided:
Calculate ΔHof of SO2(g) given the following: (answer in kJ)
S(s)+1.5O2(g)→SO3(g),ΔH=−395.8kJS_{(s)}+1.5O_{2(g)}\to SO_{3(g)}, \Delta H=-395.8kJ
2SO2(g)+O2(g)→2SO3(g),ΔH=−198.2kJ2SO_{2(g)}+O_{2(g)}\to 2SO_{3(g)}, \Delta H=-198.2kJ
Intramolecular bonding: Strength of Covalent Bonds
Calculate the enthalpy for the following reaction.
2 C2H6 (g) + 7 O=O (g) → 4 O=C=O (g) + 6 H-O-H (g)
The following are bond dissociation energies (kJ/mol):
Practice: Hydrogenation of Acetylene
Using the table below, answer the following questions about the hydrogenation reactions of acetylene:
a) Calculate the enthalpy of formation of ethylene, C2H4, considering ΔHrxn = -174.4kJ/mol
C2H2+H2→C2H4C_2H_2+H_2\to C_2H_4
Find the standard molar enthalpy of combustion of C2H5OH using the standard enthalpy of formations below.
ΔHf(C2H5OH(l))=−277.7 kJ/mol\Delta H_f(C_2H_5OH_{(l)}) = -277.7\ kJ/mol
ΔHf(CO2(g))=−393.5 kJ/mol\Delta H_f(CO_{2(g)}) = -393.5\ kJ/mol
Thermochemistry Calculations
The standard enthalpy of fusion for water, ΔHofusion = 6.01 kJ/mol. Calculate the energy required (in kJ) to convert 26.0 g of ice at 0oC to liquid water at 20oC.
Heat of Formation
Determine the standard molar enthalpy of formation of solid Al2O3 from the following equations.
Al (s) + 3 H2O (l) → Al(OH)3 (s) + 3/2 H2 (g) ΔH =+100 kJmol\Delta H\ =+100\ \frac{kJ}{mol}
H2 (g) + 1/2 O2 (g) → H2O (l)ΔH=−300 kJmol\Delta H=-300\ \frac{kJ}{mol}