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Representing Energy Changes
UNIT 3 Representing Energy Changes How can energy changes be represented in chemical reactions? Thermochemical equations with energy term as product or reactants e.g N2(g) + 2O2(g) kJ → 2NO2(g) Thermochemical equations with energy term beside the equation e.g N2(g) + 2O2(g) → 2NO2(g) ΔHr = kJ Enthalpy diagrams Image source: MHR, Chemistry 12 © ISBN ; page 294
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What to use when UNIT 3 When the problem is about:
Chapter 5: Energy Changes Section 5.2 What to use when When the problem is about: A solid dropped into water (like the copper penny expt): Use: Qreleased = Q gained where Q= mc∆T If it is as above but the container c and m is given use: Q released by reaction = Q gained by water + Q gained by container If the question is asking for molar enthalpy of reaction, and gives c and m, then calculate Q first (which is ∆H) and then use ∆H = n∆H If it is about enthalpy in aqueous solutions, use C= n/V and if as above it is a solid and molar enthalpy is needed, then you need: n=m/M
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UNIT 3 Section 5.3 Hess’s Law - calculating the enthalpy change of reactions using existing data - When using calorimetry is impractical The enthalpy change of any reaction can be determined if: the enthalpy changes of a set of reactions “add up to” the overall reaction of interest standard enthalpy change, ΔH°, values are used Image source: MHR, Chemistry 12 © ISBN ; page
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UNIT 3 Chapter 5: Energy Changes Section 5.3 For example To find the enthalpy change for formation of SO3 from O2 and S8, you can use Because only 1 mol of sulfur trioxide is in the final step: Divide the first equation by 8 so 1 mol of sulfur dioxide is in the first step Divide the second equation by 2 Image source: MHR, Chemistry 12 © ISBN ; page
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Techniques for Manipulating Equations
UNIT 3 Section 5.3 Techniques for Manipulating Equations You can reverse an equation the products become the reactants, and reactants become the products the sign of the ΔH value must be changed You can multiply each coefficient all coefficients in an equation are multiplied by the same integer or fraction the value of ΔH must also be multiplied by the same number Image source: MHR, Chemistry 12 © ISBN ; page
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Standard Molar Enthalpies of Formation
UNIT 3 Chapter 5: Energy Changes Section 5.3 Standard Molar Enthalpies of Formation Often used for Hess’ Law is standard molar enthalpy of formation, ΔH˚f the change in enthalpy when 1 mol of a compound is synthesized from its elements in their most stable form at SATP conditions enthalpies of formation for elements in their most stable state under SATP conditions are set at zero since formation equations are for 1 mol of compound, many equations include fractions (for a balanced eq’n)
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Formation Reactions and Thermal Stability
UNIT 3 Chapter 5: Energy Changes Section 5.3 Formation Reactions and Thermal Stability The thermal stability of a substance is the ability of the substance to resist decomposition when heated. decomposition is the reverse of formation the opposite sign of an enthalpy change of formation for a compound is the enthalpy change for its decomposition the greater the enthalpy change for the decomposition of a substance, the greater the thermal stability of the substance Image source: MHR, Chemistry 12 © ISBN ; page
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Using Enthalpies of Formation and Hess’s Law
UNIT 3 Chapter 5: Energy Changes Section 5.3 Using Enthalpies of Formation and Hess’s Law So the enthalpy of a reaction = the sum of all the products – the sum of all the reactants. For example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) Image source: MHR, Chemistry 12 © ISBN ; page
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C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
UNIT 3 Section 5.3 Try this Determine ∆H˚r for the following reaction using the enthalpies of formation that are provided. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ∆H˚f of C2H5OH(l): –277.6 kJ/mol ∆H˚f of CO2(g): –393.5 kJ/mol ∆H˚f of H2O(l): –285.8 kJ/mol
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UNIT 3 ∆H˚r = [(2 mol)(∆H˚f CO2(g)) + (3 mol)(∆H˚f H2O(l))] –
Chapter 5: Energy Changes Section 5.3 ∆H˚r = [(2 mol)(∆H˚f CO2(g)) + (3 mol)(∆H˚f H2O(l))] – [(1 mol)(∆H˚f C2H5OH(l)) + (3 mol)(∆H˚fO2(g)] ∆H˚r = [(2 mol)(–393.5 kJ/mol) + (3 mol)(–285.8 kJ/mol)] – [(1 mol)(–277.6 kJ/mol) + (3 mol)(0 kJ/mol)] Image source: MHR, Chemistry 12 © ISBN ; page 98 ∆H˚r = (– kJ) – (–277.6 kJ) ∆H˚r = – kJ
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