Energy and Chemical Change

A lead pellet having a mass of 26. 47 g at 89 A lead pellet having a mass of 26.47 g at 89.98 oC was placed in a constant pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50 oC to 23.17 oC. What is the specific heat of the lead pellet? Note: treat the calorimeter as an isolated system, the heat gained by the water is equal to the heat lost by the lead pellet. There is enough information to calculate the heat gained by the water. If you know the heat gained by the water, you also know the heat lost by the Pb pellet. qH2O = msΔT = (100.0 g)(4.184 J/goC)(23.17-22.50 oC) = 280.3 J So qPb= -280.3 J Now there is enough information to use the heat equation to solve for the specific heat (s) of Pb. sPb= qPb/(mΔT) = (-280.3 J)/(26.47 g x (23.17-89.98 oC)) = 0.1584 J/g oC 4/22/2019 S.A. McFarland©2006

Heats of reaction The amount of heat that a reaction produces or absorbs depends on the number of moles of reactant that react A set of standard states have been defined for reporting heats of reactions 1 atm pressure for all gases 1 M concentration for aqueous solutions temperature of 25 °C (298 K) is often specified as well The standard heat of reaction is the value of the enthalpy change occurring under standard conditions involving the actual number of moles specified by the equation coefficients An enthalpy change for standard conditions is denoted ΔHo For example, the thermochemical equation for the production of ammonia from it elements at standard conditions is: The law of conservation of energy requires 4/22/2019 S.A. McFarland©2006

Enthalpy is a state function Two paths for the formation of carbon dioxide gas; each give the same enthalpy change Enthalpy changes for reactions can be calculated by algebraic summation Hess’s Law: The value of the enthalpy change for any reaction that can be written in steps equals the sum of the values of the enthalpy change of each of the individual steps Enthalpy changes for a huge number of reactions may be calculated using only a few simple rules 4/22/2019 S.A. McFarland©2006

Rules for manipulating thermochemical equations When an equation is reversed the sign of the enthalpy change must also be reversed. Formulas canceled from both sides of an equation must be for substances in identical physical states. If all the coefficients of an equation are multiplied or divided by the same factor, the value of the enthalpy change must likewise be multiplied or divided by that factor. 4/22/2019 S.A. McFarland©2006

Because there is no way to measure the absolute value of the enthalpy of a substance, we measure changes in enthalpy that occur during a reaction. There is no need to make the measurement for every single reaction of interest if we remember Hess’ Law. Establish an arbitrary scale with the standard enthalpy of formation (ΔH0f) as a reference point for all enthalpy expressions. Standard enthalpy of formation (ΔH0f) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. The standard enthalpy of formation of any element in its most stable form is zero. See Table 6.4. ΔH0 (C, graphite) = 0 f ΔH0 (O2) = 0 f ΔH0 (C, diamond) = 1.90 kJ/mol f ΔH0 (O3) = 142 kJ/mol f 4/22/2019 S.A. McFarland©2006

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The standard enthalpy of reaction (ΔH0 ) is the enthalpy of a reaction carried out at 1 atm. rxn aA + bB cC + dD ΔH0 rxn dΔH0 (D) f cΔH0 (C) = [ + ] - bΔH0 (B) aΔH0 (A) = ΔH0 rxn nΔH0 (products) f Σ [ mΔH0 (reactants)] Σ[ ] - 4/22/2019 S.A. McFarland©2006

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l) DH0 rxn nDH0 (products) f = S mDH0 (reactants) - DH0 rxn 6DH0 (H2O) f 12DH0 (CO2) = [ + ] - 2DH0 (C6H6) DH0 rxn = [ 12(-393.5) + 6(-285.8 )] – [ 2(49.04) ] = -6534 kJ -6534 kJ 2 mol = - 3267 kJ/mol C6H6 4/22/2019 S.A. McFarland©2006

Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ rxn S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJ rxn CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ rxn 1. Write the enthalpy of formation reaction for CS2 C(graphite) + 2S(rhombic) CS2 (l) 2. Add the given rxns so that the result is the desired rxn. rxn C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ 2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJ rxn + CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJ rxn C(graphite) + 2S(rhombic) CS2 (l) DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ rxn 4/22/2019 S.A. McFarland©2006

DHsoln = Hsoln - Hcomponents The enthalpy of solution (ΔHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. DHsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 4/22/2019 S.A. McFarland©2006

The Electronic Structure of Atoms Ch. 7

Properties of Waves Wavelength (λ) is the distance between identical points on successive waves. Amplitude is the vertical distance from the midline of a wave to the peak or trough. 4/22/2019 S.A. McFarland©2006

The speed (u) of the wave = λ x ν Properties of Waves Frequency (v) is the number of waves that pass through a particular point in 1 second (Hz = 1 cycle/s). The speed (u) of the wave = λ x ν 4/22/2019 S.A. McFarland©2006

Speed of light (c) in vacuum = 3.00 x 108 m/s Maxwell (1873), proposed that visible light consists of electromagnetic waves. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves. Speed of light (c) in vacuum = 3.00 x 108 m/s All electromagnetic radiation λ x ν = c 4/22/2019 S.A. McFarland©2006