Thermochemical Equations

Chemistry 11(C)

Lesson Objectives• Classify reactions as endothermic or

exothermic• Complete calculations using thermochemical

equations

Thermochemical Equations

Endothermic and Exothermic Reactions

• Energy can be absorbed or released

• Exothermic – energy is released from a reaction

• Endothermic – energy is absorbed by a reaction

• Energy is released when bonds form

• Energy is absorbed to break bonds

Releasing and Absorbing Energy

Ex) 2H2+ O2 2H2O

Reactants

Products

Energy absorbed

Energy released

• Exothermic ⇒ less E is absorbed to break bonds than is released when bonds form

Bond Energy reactant < Bond

Energy product

• Endothermic ⇒ more E is absorbed to break bonds than is released when bonds form

Bond Energy reactant > Bond

Energy product

Classifying Reactions

• Enthalpy change – (∆H); amount of energy released or absorbed as heat by a system when the pressure is constant

• ∆Hrxn = Hproducts – Hreactants

– Exothermic⇒ ∆H = negative values– Endothermic⇒ ∆H = positive values

Enthalpy

Hreactants > Hproducts

Exothermic

Enth

alpy

Reaction Progress

Hreactants < Hproducts

Endothermic

Enth

alpy

Reaction Progress

• Thermochemical equation – chemical equation that includes the enthalpy change

– Coefficients represent the number of moles– ∆H is directly proportional to the number of

moles– Include physical states

Thermochemical Equations

Ex) 4Fe(s) + 3O2(g)2Fe2O3(s) + 1625 kJ or

4Fe(s) + 3O2(g)2Fe2O3(s) ∆H= –1625 kJ

• Standard enthalpy values for specific chemicals can be found in reference tables

• Phase of the chemical will affect the enthalpy value – Elements in their

natural state will have ∆H0

f of zero

Enthalpy Values

Substance ∆H0f

(kJ/mol) BaCl2(s) –855.0Ca(s) 0CaCl2(s) –795.4Br2(l ) 0CO(g) –110.5CO2(g) –393.5 CuO(s) –157.3 Fe2O3(s) –824.2H2(g) 0H2O(g) –241.8H2O(l ) –285.8

• Enthalpy of reaction equals the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants

Calculating ∆Hrxn from ∆H0f

Substance

∆H0f

(kJ/mol) CH4(g) –74.6Cl2(g) 0CCl4(l ) –139H2(g) 0

Ex) CH4(g) + 2Cl2(g) CCl4(l ) + 2H2(g)

∆Hrxn= ∆Hf0 (products) - ∆Hf

0 (reactants)

∆Hrxn= (–139 + (2×0))–(–74.6 + (2×0))

∆Hrxn= –64.4 kJ/mol

• Hess’s Law – the sum of the enthalpy changes for each step of a reaction is equal to the overall enthalpy change– Manipulate the equations so the sum of the given

equations equals the overall equation• If the reaction is reversed, the sign of ∆H must be

reversed

• Multiply given equations by coefficients

Hess’s Law

Ex) 4Fe(s) + 3O2(g)2Fe2O3(s) ∆H= –1625 kJ 2Fe2O3(s)4Fe(s) + 3O2(g) ∆H= 1625 kJ

3[ 2Fe2O3(s)4Fe(s) + 3O2(g)∆H= 1625 kJ ] 6Fe2O3(s)12Fe(s) + 9O2(g) ∆H= 4875 kJ

Ex) 2H2O2(l ) → 2H2O(l ) + O2(g)∆H=?

2H2(g) + O2(g) → 2H2O(l )

H2(g) + O2(g) → H2O2(l ) ∆H1= –188 kJ

∆H2= –572 kJ

1. Rearrange formulas so each chemical is on the same side of the reaction as the goal formula

∆H= –196 kJ

H2O2(l) → H2(g) + O2(g) 188 kJ

3. Find the sum of the thermochemical equations

2. Multiply by coefficients so the number of moles of each chemical match the goal formula- Keep in mind that matching chemicals on

opposite sides of the combined reaction will cancel each other out

2H2O2(l ) → 2H2(g) + 2O2(g) 376 kJ

+

Hess’s Law

Lesson Objectives• Classify reactions as endothermic or

exothermic• Complete calculations using thermochemical

equations– Solve for heat of reaction when given heats of

formation– Solve problems using Hess’s law

Thermochemical Equations