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Thermochemical Equations

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Thermochemical Equations

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Chemistry 11(C)

- 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 enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactantsLaw – 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 enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants

- 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 enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants

- 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