CHAPTER 10 Chemical Reactions 10.4 Chemical Reactions and Energy

There are three key components to a chemical reaction: Reactants Products Energy (in or out)

Chapter 4.2 Chemical Reactions There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions

Chapter 4.2 Chemical Reactions There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Photosynthesis is an endothermic reaction: energy is absorbed

Chapter 4.2 Chemical Reactions There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Photosynthesis is an endothermic reaction: energy is absorbed Cellular respiration is an exothermic reaction: energy is released

Properties of Solutions There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions

Energy is absorbed from the surroundings so the pack feels cold There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Energy is absorbed from the surroundings so the pack feels cold

Energy is released into the surroundings so the pack feels hot There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Energy is released into the surroundings so the pack feels hot

Properties of Solutions There are three key components to a chemical reaction: Reactants Products Energy (in or out) Chapter 4.2 Chemical Reactions Chapter 9.3 Properties of Solutions Change in enthalpy enthalpy: the amount of energy that is released or absorbed during a chemical reaction

Enthalpy change (∆H, J/mole) Reaction Exothermic Endothermic Energy is released is absorbed Enthalpy change (∆H, J/mole) is a negative number ∆H < 0 is a positive number ∆H > 0

Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ

Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ Reactants Products Energy thermochemical equation: the equation that gives the chemical reaction and the energy information of the reaction.

Enthalpy calculations Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ The reverse chemical reaction involves the same amount of energy, but the energy flow is reversed (“in” instead of “out”): CO2(g) C(s) + O2(g) ∆H = +393.5 kJ

Enthalpy calculations Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ 1 mole 1 mole 1 mole The combustion of twice as much carbon releases twice as much energy: 2C(s) + 2O2(g) 2CO2(g) ∆H = –787.0 kJ 2 moles 2 moles 2 moles

Enthalpy calculations Chemical equation for the formation of rust: 2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ 2 moles 3/2 moles 1 mole

Enthalpy calculations Chemical equation for the formation of rust: 2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ 2 moles 3/2 moles 1 mole Rewrite the chemical equation using coefficients with the smallest whole numbers possible

Enthalpy calculations Chemical equation for the formation of rust: 2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ 2 moles 3/2 moles 1 mole x 2 4Fe(s) + 3O2(g) 2Fe2O3(s) ∆H = ? 4 moles 3 moles 2 moles What is the enthalpy change for this reaction?

Enthalpy calculations Chemical equation for the formation of rust: 2Fe(s) + 3/2O2(g) Fe2O3(s) ∆H = –824.2 kJ 2 moles 3/2 moles 1 mole x 2 x 2 4Fe(s) + 3O2(g) 2Fe2O3(s) ∆H = –1,648.4 kJ 4 moles 3 moles 2 moles

Enthalpy of formation Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ This is also the chemical equation for the formation of CO2. ∆Hreaction = ∆Hformation of CO2 = –393.5 kJ

Enthalpy of formation Chemical equation for the combustion of carbon: C(s) + O2(g) CO2(g) ∆H = –393.5 kJ This is also the chemical equation for the formation of CO2. ∆Hreaction = ∆Hformation of CO2 = –393.5 kJ ∆Hf (CO2) = –393.5 kJ/mole The formation of 1 mole of CO2 releases 393.5 kJ of energy

Enthalpy of formation Enthalpies of formation of some common substances

Enthalpy of formation Enthalpies of formation of some common substances Knowing these values and the following equation, you can calculate unknown enthalpy values:

Enthalpy of formation Let’s see an example! Enthalpies of formation of some common substances Let’s see an example! Knowing these values and the following equation, you can calculate unknown enthalpy values:

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose.

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. Asked: ∆Hf(glucose) = ?

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. Asked: ∆Hf(glucose) = ? Given: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJ From the table of enthalpies of formation Note: In this problem we have to use the enthalpy of formation of gaseous water which is different from the enthalpy of formation of liquid water. In the text solved problem 313 the wrong enthalpy of formation is used.

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. Asked: ∆Hf(glucose) = ? Given: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJ From the table of enthalpies of formation Relationships:

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJ Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ∆H = –2,808 kJ Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products Relationships: 10.4 Chemical Reactions and Energy

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products Remember to multiply by the coefficients!

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ reactants products

Enthalpy calculations The complete combustion of glucose (C6H12O6) releases 2,808 kJ per mole of glucose. Calculate the enthalpy of formation of glucose. Formation of glucose 6CO2(g) + 6H2O(g) C6H12O6(s) + 6O2(g) ∆H = +2,808 kJ 1 mole Asked: ∆Hf(glucose) = ? Answer: ∆Hf(glucose) = –1,004 kJ/mole

A B A B + ∆H = X

The reverse reaction changes the sign of ∆H B A B + ∆H = X A B A B + ∆H = –X The reverse reaction changes the sign of ∆H

If three times more substances are involved, ∆H is three times greater + ∆H = X x 3 A B A B ∆H = 3X + If three times more substances are involved, ∆H is three times greater

∆H(reaction) = ∆Hf (products) – ∆Hf (reactants) B A B + ∆H = X ∆H(reaction) = ∆Hf (products) – ∆Hf (reactants) A B A B ∆H(reaction) = ∆Hf – ∆Hf + ∆Hf

Energy profile Thermochemical equation: Reactants Products ∆H = … kJ

also have stored energy Energy profile Thermochemical equation: Reactants Products ∆H = … kJ have stored energy also have stored energy Energy flow during the reaction

We can graph the change in energy as the reaction takes place Energy profile Thermochemical equation: Reactants Products ∆H = … kJ have stored energy also have stored energy Energy flow during the reaction We can graph the change in energy as the reaction takes place

We can graph the change in energy as the reaction takes place Energy profile Thermochemical equation: Reactants Products ∆H = … kJ have stored energy also have stored energy Energy flow during the reaction Energy We can graph the change in energy as the reaction takes place Progress of reaction

also have stored energy Energy profile Thermochemical equation: Reactants Products ∆H = … kJ have stored energy also have stored energy Energy flow during the reaction

The reaction cannot start without this initial input of energy Energy profile Activation energy The reaction cannot start without this initial input of energy

Energy profile Combustion of carbon: C(s) + O2(g) CO2(g) Wood does not spontaneously light itself up on fire

Energy profile Reaction of sodium in water: Na(s) + H2O(l) 2NaOH + H2(g) Sodium reacts with water immediately (and violently) upon contact

Hess’s law Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2).

Hess’s law Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A ∆H1

Hess’s law Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A ∆H1 A B ∆H2

Hess’s law Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A ∆H1 A B ∆H2 B P ∆H3

Hess’s law Hess’s law: the overall enthalpy of a reaction (1) is the sum of the reaction enthalpies of the various steps into which a reaction can be divided (2). R A ∆H1 A B ∆H2 B P ∆H3 R P ∆H4

Hess’s law ∆H1 ∆H2 ∆H3 ∆H4 Hess’s law: ∆H4 = ∆H1 + ∆H2 + ∆H3 R A A B B P ∆H3 R P ∆H4

Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite.

Hess’s law Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite. Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Relationships: Hess’s law

Hess’s law Given that the enthalpy of combustion for graphite (Cgr) and diamond (Cd) are –393.5 kJ/mole and –395.4 kJ/mole, respectively, calculate the enthalpy of formation of diamond from graphite. Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Relationships: Hess’s law Strategy: Create a path that leads from Cgr to Cd.

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Cgr(s) is a reactant in: Cgr(s) Cd(s)

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Cgr(s) is a reactant in: Cgr(s) Cd(s) Cgr(s) is also a reactant in: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Cd(s) is a product in: Cgr(s) Cd(s)

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Cd(s) is a product in: Cgr(s) Cd(s) Cd(s) is a reactant in: Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Cd(s) is a product in: Cgr(s) Cd(s) Cd(s) is a reactant in: Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the reverse reaction so that Cd(s) is a product, and adjust DH: CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the sum of the two equations: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the sum of the two equations: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the sum of the two equations: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the sum of the two equations: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole Cgr(s) Cd(s) ∆H = (–395.4 + 395.4) kJ/mole

Hess’s law Asked: Cgr(s) Cd(s) ∆H = ? Given: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole Cd(s) + O2(g) CO2(g) ∆H = –395.4 kJ/mole Write the sum of the two equations: Cgr(s) + O2(g) CO2(g) ∆H = –393.5 kJ/mole CO2(g) Cd(s) + O2(g) ∆H = +395.4 kJ/mole Cgr(s) Cd(s) ∆H = +1.9 kJ/mole

Energy profile of a reaction Hess’s law