8. Thermochemistry

Hess's Law

Enthalpy of Reaction

ALEKS - Using Hess's Law to Calculate Net Reaction Enthalpy

This reaction has an equilibrium constant of Kp = 2.26 * 104 at 298 K.
CO(g) + 2 H2(g) ⇌ CH3OH(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. a. CH3OH(g) ⇌ CO(g) + 2 H2(g)

Given the data
N21g2 + O21g2¡ 2 NO1g2 H = +180.7 kJ
2 NO1g2 + O21g2¡ 2 NO21g2 H = -113.1 kJ
2 N2O1g2¡ 2 N21g2 + O21g2 H = -163.2 kJ
use Hess’s law to calculate H for the reaction
N2O1g2 + NO21g2¡ 3 NO1g2

The industrial degreasing solvent methylene chloride, CH2Cl2, is prepared from methane by reaction with chlorine: CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) Use the following data to calcualte ΔH° in kilojoules for the reaction: CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ΔH° = -98.3 kJ CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ΔH° = -104 kJ

The reaction 2 AsH31g2 ∆ As21g2 + 3 H21g2 has Kp =
7.2 * 107 at 1073 K. At the same temperature, what is Kp
for each of the following reactions?
(a) As21g2 + 3 H21g2 ∆ 2 AsH31g2

Consider the combustion of liquid methanol, CH3OH1l2: 
CH3OH1l2 + 32O21g2¡ CO21g2 + 2 H2O1l2 H = -726.5 kJ
(b) Balance the forward reaction with whole-number coefficients.
What is H for the reaction represented by this
equation?

Show how Equations 18.7 and 18.9 can be added to give Equation 18.10.

This reaction has an equilibrium constant of Kp = 2.2 * 106 at 298 K.
2 COF2(g) ⇌ CO2(g) + CF4(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. b. 6 COF2(g) ⇌ 3 CO2(g) + 3 CF4(g)

The ΔH °f of TiI3(s) is -328 kJ>mol and the ΔH ° for the reaction
2 Ti(s) + 3 I2( g)¡2 TiI3(s) is -839 kJ. Calculate the ΔH of
sublimation of I2(s), which is a solid at 25 °C.

The following steps occur in the reaction of ethyl alcohol (CH3CH2OH) wiht oxygen to yield acetic acid (CH3CO2H). Show that equations 1 and 2 sum to give the net equation and calculate ΔH° for the net equation. (1) CH3CH2OH(l) + 1/2 O2(g) → CH3CHO (g) + H2O(l) ΔH° = -174.2 kJ (2) CH3CHO(g) + 1/2 O2(g) → CH3CO2H(l) ΔH° = -318.4 kJ (Net) CH3CH2OH(l) + O2(g) → CH3CO2H(l) + H2O(l) ΔH° = ?

Set up a Hess's law cycle, and use the following information to calculate ΔH°f for aqueous nitiric acid, HNO3(aq). You will need to use fractional coefficients for some equations. 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) ΔH° = -137.3 kJ 2 NO(g) + O2(g) → 2 NO2(g) ΔH° = -116.2 kJ 4 NH3(g) + 5 O2(g) → 4 NO (g) + 6 H2O(l) ΔH° = -1165.2 kJ NH3(g) ΔH°f = -46.1 kJ/mol H2O(l) ΔH°f = -285.8 kJ/mol

Consider the following equilibrium, for which
Kp = 0.0752 at 480 C:
2 Cl21g2 + 2 H2O1g2 Δ 4 HCl1g2 + O21g2
(a) What is the value of Kp for the reaction
4 HCl1g2 + O21g2 Δ 2 Cl21g2 + 2 H2O1g2?

This reaction has an equilibrium constant of Kp = 2.2 * 106 at 298 K.
2 COF2(g) ⇌ CO2(g) + CF4(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. c. 2 CO2(g) + 2 CF4(g) ⇌ 4 COF2(g)

 Correct calculation and manipulation of the equations. 

One of the reactions has to be reversed and this changes of the sign of &#916;H.

One of the reactions has to be reversed and this changes of the sign of &#916;H and the given reaction contains two moles of NO2.

If a chemical reaction is expressed as the sum of a series of steps, then &#916;Hrxn for the overall equation is the sum of the heats of reaction for each step.

Find &#916;Hrxn for the following reaction: N2 + 2 O2 &#8594; 2 NO2 Based on the following data. 2 NO &#8594; N2 + O2&nbsp;&nbsp;&nbsp;&nbsp;&#916;H = &#x2212;180 kJ 2 NO + O2 &#8594; 2 NO2&nbsp;&nbsp;&nbsp;&nbsp;&#916;H = &#x2212;112 kJ

Calculate ΔHrxn for the reaction: 
5 C(s) + 6 H2( g)¡C5H12(l ) 
Use the following reactions and given ΔH’s: 
C5H12(l ) + 8 O2( g)¡5 CO2( g) + 6 H2O( g) 
ΔH = -3244.8 kJ 
C(s) + O2( g)¡CO2( g) ΔH = -393.5 kJ 
2 H2( g) + O2( g)¡2 H2O( g) ΔH = -483.5 kJ

For each generic reaction, determine the value of ΔH2 in terms
of ΔH1.
a. A + B¡2 C ΔH1
2 C¡A + B ΔH2 = ?

Practice Problem: Hess's Law

Calculate the enthalpy change for the reaction
C(s) + 2 H2(g) → CH4(g) ΔH = ?
Given the enthalpy values for the following reactions
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = −890.4 kJ
C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ
H2(g) + 1/2 O2(g) → H2O (g) ΔH = −285.8 kJ
(a) −1569.7 kJ
(b) +211.1 kJ
(c) −1855.5 kJ
(d) −74.7 kJ

The reaction of A with B to give D proceeds in two steps:
(1) A + B → C ΔH° = -20 kJ
(2) C + B → D ΔH° = +50 kJ
(3) A + 2B → D ΔH° = ?
(b) What is the value of ΔH° for the overall reaction
A + 2 B → D ΔH° = ?

Consider the following equilibrium:
2 H21g2 + S21g2 Δ 2 H2S1g2 Kc = 1.08 * 107at 700 C
(c) Calculate the value of Kc
if you rewrote the equation H21g2 + 12S21g2 Δ H2S1g2.

Multiplication of the original reaction results in multiplication of ∆Hrxn.Division results in the division of ∆Hrxn..Reversing the original reaction results in reversing the sign of ∆Hrxn.

Addition of ∆Hrxn values of multiple steps gives us the ∆Hrxn of the overall reaction.

For each generic reaction, determine the value of ΔH2 in terms
of ΔH1.
b. A + 12 B¡C ΔH1
2 A + B¡2 C ΔH2 = ?

9.149 Consider the reaction: 4 CO1g2 2 NO21g2 4 CO21g2 N21g2. Using the following information, determine ΔH° for the reaction at 25 °C. 
NO1g2 ΔH°f = + 91.3 kJ>mol 
CO21g2 ΔH°f = - 393.5 kJ>mol 2 NO1g2 + O21g2 S 2 NO21g2 ΔH° = - 116.2 kJ 
2 CO1g2 + O21g2 S 2 CO21g2 ΔH° = - 566.0 kJ

Consider the decomposition of liquid benzene, C6H61l2, to
gaseous acetylene, C2H21g2:
C6H61l2 ¡ 3 C2H21g2 H = +630 kJ
(a) What is the enthalpy change for the reverse reaction?

Calculate the&nbsp;∆Hrxn&nbsp;for&nbsp;<img src="https://static.studychannel-prd.pearsonprd.tech/courses/general-chemistry/thumbnails/images/7e9d8831"class="fr-fic fr-dii">Given the following set of reactions:&nbsp; <img src="https://lightcat-files.s3.amazonaws.com/problem_images/42ed3b932dd03fb5-1608574750112.jpg"class="fr-fic fr-dii">

This reaction has an equilibrium constant of Kp = 2.2 * 106 at 298 K.
2 COF2(g) ⇌ CO2(g) + CF4(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. a. COF2 (g) ⇌ 1/2 CO2(g) + 1/2 CF4(g)

Relationships Involving ΔHrxn - Hess's Law

Hess’s law can be used to calculate reaction enthalpies for hypothetical processes that can’t be carried out in the labo- ratory. Set up a Hess’s law cycle that will let you calculate ∆H° for the conversion of methane to ethylene: 2 CH4(g) → C2H4(g) + 2 H2(g) You can use the following information: 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l) ∆H° = -3120.8 kJ CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H° = -890.3 kJ C2H4(g) + H2(g) → C2H6(g) ∆H° = -136.3 kJ H2O(l) ∆H°f = -285.8 kJ/mol

From the enthalpies of reaction
H21g2 + F21g2¡ 2 HF1g2 H = -537 kJ
C1s2 + 2 F21g2¡ CF41g2 H = -680 kJ
2 C1s2 + 2 H21g2¡ C2H41g2 H = +52.3 kJ
calculate H for the reaction of ethylene with F2:
C2H41g2 + 6 F21g2¡ 2 CF41g2 + 4 HF1g2

Calculate ΔHrxn for the reaction:
Fe2O3(s) + 3 CO( g)¡2 Fe(s) + 3 CO2( g)
Use the following reactions and given ΔH’s:
2 Fe(s) + 32 O2( g)¡Fe2O3(s) ΔH = -824.2 kJ
CO( g) + 12 O2( g)¡CO2( g) ΔH = -282.7 kJ

Hess's Law Common Test Question

Consider the generic reaction:
A + 2 B¡C + 3 D ΔH = 155 kJ
Determine the value of ΔH for each related reaction.
a. 3 A + 6 B¡3 C + 9 D
b. C + 3 D¡A + 2 B
c. 12 C + 32 D¡12 A + B

Consider the following hypothetical reactions:
A ¡B HI = +60 kJ
B ¡ C HII = -90 kJ
(b) Construct an enthalpy diagram for substances A, B, and
C, and show how Hess’s law applies.

This reaction has an equilibrium constant of Kp = 2.26 * 104 at 298 K.
CO(g) + 2 H2(g) ⇌ CH3OH(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. c. 2 CH3OH(g) ⇌ 2 CO(g) + 4 H2(g)

The reaction of A with B to give D proceeds in two steps:
(1) A + B → C ΔH° = -20 kJ
(2) C + B → D ΔH° = +50 kJ
(3) A + 2B → D ΔH° = ?
(a) Which Hess’s law diagram represents the reaction steps and the overall reaction?
Diagram 1
Diagram 2

Hess’s Law

This reaction has an equilibrium constant of Kp = 2.26 * 104 at 298 K.
CO(g) + 2 H2(g) ⇌ CH3OH(g)
Calculate Kp for each reaction and predict whether reactants or
products will be favored at equilibrium. b. 1/2 CO(g) + H2 (g) ⇌ 1/2 CH3OH(g)

For each generic reaction, determine the value of ΔH2 in terms 
of ΔH1. 
c. A¡B + 2 C ΔH1 
12 B + C¡12 A ΔH2 = ?

Hess's Law Example 1

Hess's Law Example 2

Calculate the&nbsp;∆Hrxn for&nbsp;<img src="https://static.studychannel-prd.pearsonprd.tech/courses/general-chemistry/thumbnails/images/bf5019cc"class="fr-fic fr-dii">Given the following reactions:<img src="https://lightcat-files.s3.amazonaws.com/problem_images/8f2ec113a0c2cc5a-1608574782596.jpg"class="fr-fic fr-dii">

Acetylene, C2H2, reacts with H2 in two steps to yield ethane, CH3CH3:
(1) HC?CH + H2 → H2C?CH2 ΔH° = -175.1 kJ
(2) H2C?CH2 + H2 → CH3CH3 ΔH° = -136.3 kJ
Which arrow (a–c) in the Hess’s law diagram corresponds 
to which step, an2d which a2rrow cor2responds to the net reaction? Where are the reactants located on the diagram, and where are the products located?
<img src="https://lightcat-files.s3.amazonaws.com/problem_images/79f72c7ab9b1c632-1669905413233.jpg" alt="" />

Sulfuric acid (H2SO4), the most widely produced chemical in the world, is amde yb a two-step oxidaiton of sulfur to sulfur trioxide, SO3, followed by reaciton with water. Calculate ΔH°f for SO3 in kJ/mol, given the following data: S(s) + O2(g) → SO2(g) ΔH° = -296.8 kJ SO2(g) + 1/2 O2(g) → SO3(g) ΔH° = -98.9 kJ

Calculate the enthalpy change for the reaction
P4O61s2 + 2 O21g2 ¡ P4O101s2
given the following enthalpies of reaction:
P41s2 + 3 O21g2 ¡P4O61s2 H = -1640.1 kJ
P41s2 + 5 O21g2 ¡P4O101s2 H = -2940.1 kJ

Consider the combustion of liquid methanol, CH3OH1l2:
CH3OH1l2 + 32O21g2¡ CO21g2 + 2 H2O1l2 H = -726.5 kJ
(a) What is the enthalpy change for the reverse reaction?

Hess's Law Trick Question You Should Know

Hess's Law and Heats of Formation

From the enthalpies of reaction
2 C1s2 + O21g2¡ 2 CO1g2 H = -221.0 kJ
2 C1s2 + O21g2 + 4 H21g2¡ 2 CH3OH1g2 H = -402.4 kJ
calculate H for the reaction
CO1g2 + 2 H21g2¡CH3OH1g2

Calculate the&nbsp;∆Hrxn&nbsp;for the following thermochemical equation:&nbsp;<img src="https://static.studychannel-prd.pearsonprd.tech/courses/general-chemistry/thumbnails/images/db18c05c"class="fr-fic fr-dii">When given the following:&nbsp; <img src="https://lightcat-files.s3.amazonaws.com/problem_images/ae13fba2d635c857-1608574706932.jpg"class="fr-fic fr-dii">

Hydrazine, a component of rocket fuel, undergoes combus- tion to yield N2 and H2O:
N2H41l2 + O21g2 S N21g2 + 2 H2O1l2 (b) Use the following information to set up a Hess’s law cycle, and then calculate ΔH° for the combustion reac- tion. You will need to use fractional coefficients for some equations. 
2 NH31g2 + 3 N2O1g2 S 4 N21g2 + 3 H2O1l2 
ΔH° = - 1011.2 kJ N2O1g2 + 3 H21g2 S N2H41l2 + H2O1l2 
ΔH° = - 317.2 kJ 4 NH31g2 + O21g2 S 2 N2H41l2 + 2 H2O1l2 
ΔH° = - 286.0 kJ 
H2O1l2 ΔH°f = - 285.8 kJ>mol

Calculate ΔHrxn for the reaction: 
CH4( g) + 4 Cl2( g)¡CCl4( g) + 4 HCl( g) 
Use the following reactions and given ΔH’s: 
C(s) + 2 H2( g)¡CH4( g) ΔH = -74.6 kJ 
C(s) + 2 Cl2( g)¡CCl4( g) ΔH = -95.7 kJ 
H2( g) + Cl2( g)¡2 HCl( g) ΔH = -92.3 kJ

Find ∆H° in kilojoules for the reaciton of nitric oxide with oxygen, 2 NO(g) + O2(g) → N2O4(g), given the following data: N2O4(g) → 2 NO2(g) ∆H° = 55.3 kJ NO(g) + 1/2 O2(g) → NO2(g) ∆H° = -58.1 kJ

What is Hess’s law, and why does it “work”?

The standard enthalpy change for the reaciton of SO3(g) with H2O(l) to yield H2SO4(aq) is ΔH° = -227.8 kJ. Use the information in Problem 9.104 to calculate ΔH°f for H2SO4(aq) in kJ/mol. [For H2O(l), ΔH°f = -285.88 kJ/mol.]

Reaction of gaseous fluorine with compound X yields a sin- gle product Y, whose mass percent composition is 61.7% F and 38.3% Cl. (c) Calculate ΔH° for the synthesis of Y using the following information:
2 CIF1g2 + O21g2 S Cl2O1g2 + OF21g2
ΔH° = + 205.4 kJ 2 CIF31l2 + 2 O21g2 S Cl2O1g2 + 3 OF21g2
ΔH° = + 532.8 kJ
OF21g2 ΔH°f = + 24.5 kJ>mol

Calculate ΔHrxn for the reaction: 
CaO(s) + CO2( g)¡CaCO3(s) 
Use the following reactions and given ΔH’s: 
Ca(s) + CO2( g) + 12 O2( g)¡CaCO3(s) ΔH = -812.8 kJ 
2 Ca(s) + O2( g)¡2 CaO(s) ΔH = -1269.8 kJ

We can use Hess’s law to calculate enthalpy changes that
cannot be measured. One such reaction is the conversion of
methane to ethane:
2 CH41g2 ¡C2H61g2 + H21g2
Calculate the H° for this reaction using the following thermochemical
data:
CH41g2 + 2 O21g2¡CO21g2 + 2 H2O1l2 H° = -890.3 kJ
2 H21g2 + O21g2¡2 H2O1l2 H ° = -571.6 kJ
2 C2H61g2 + 7 O21g2¡4 CO21g2 + 6 H2O1l2 H° = -3120.8 kJ

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[BOOK CHEM] Clutch Version

Hess's Law involves the use of partial reactions in the determination of the overall enthalpy of reaction.

Nature of Energy

Thermochemistry is the study of matter and energy associated with chemical reactions or physical changes.&nbsp;

Endothermic & Exothermic Reactions

Endothermic and Exothermic Reactions involve the absorbing and releasing of thermal energy respectively.&nbsp;&nbsp;

Heat Capacity

Heat Capacity is the amount of heat required to change the temperature of a substance.&nbsp;

Thermochemical Equations

Thermochemical Equations&nbsp;involve a balanced chemical equation with a given enthalpy value.

Internal Energy

The Internal Energy of a system is the total energy from all forms of kinetic and potential energy.&nbsp;

Kinetic & Potential Energy

Mechanical Energy is the energy associated with an object's velocity (Kinetic Energy) and its position (Potential Energy).&nbsp;

Formation Equations

Formation Equations involve the standard states of elements combining to form 1 mole of product.

Enthalpy of Formation

The Enthalpy of Formation for an element is a key component in determining the enthalpy of reaction.&nbsp;

Thermal Equilibrium

Thermal Equilibrium involves two substances that are in physical contact reaching the same final temperature over time.&nbsp;

Constant-Pressure Calorimetry

Constant-Pressure Calorimetry uses a coffee cup calorimeter to determine the heat transfers occurring in a liquid solution.

Constant-Volume Calorimetry

Constant-Volume Calorimetry uses a bomb calorimeter and a combustion reaction to determine its enthalpy of reaction.&nbsp;

First Law of Thermodynamics

The First Law of Thermodynamics states that energy cannot be created nor destroyed, but instead is transferred.&nbsp;