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Thermochemical Equations

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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

- Endothermic – energy is absorbed by a reaction

- Exothermic– energy is released from a reaction

- Energy is released when bonds form

- Energy is absorbed to break bonds

- Releasing and Absorbing Energy

Ex) 2H2+ O2 2H2O

Energy absorbed

Energy released

Reactants

Products

- 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

Exothermic

Endothermic

Enthalpy

Enthalpy

Reaction Progress

Reaction Progress

Hreactants> Hproducts

Hreactants< Hproducts

- 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 ∆H0f of zero

- Enthalpy Values

- 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

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

(–74.6 + (2×0))

∆Hrxn=

(–139 + (2×0))

–

∆Hrxn= ∆Hf0 (products) - ∆Hf0 (reactants)

∆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

- Manipulate the equations so the sum of the given equations equals the overall equation

- 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

- Hess’s Law

- 376 kJ

- 188 kJ

2H2O2(l ) → 2H2(g) + 2O2(g)

H2O2(l) → H2(g) + O2(g)

Ex) 2H2O2(l ) → 2H2O(l ) + O2(g)

∆H=?

+

H2(g) + O2(g) → H2O2(l )

∆H1= –188 kJ

∆H= –196 kJ

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

2H2(g) + O2(g) → 2H2O(l )

∆H2= –572 kJ

- 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

Find the sum of the thermochemical equations

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