Transcript Slide 1
Energy and Chemical Change Ch. 6 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 4/29/2020 S.A. McFarland©2006 2 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 • 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: – – – 1 atm pressure for all gases 1 M concentration for aqueous solutions temperature of 25 °C (298 K) is often specified as well N 2 ( g ) 3H 2 ( g ) 2NH3 ( g ) • H 92.38 kJ The law of conservation of energy requires 2NH3 ( g ) N 2 ( g ) 3H 2 ( g ) 4/29/2020 S.A. McFarland©2006 H 92.38 kJ 3 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 4/29/2020 S.A. McFarland©2006 4 Rules for manipulating thermochemical equations 1) When an equation is reversed the sign of the enthalpy change must also be reversed. 2) Formulas canceled from both sides of an equation must be for substances in identical physical states. 3) 4/29/2020 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. S.A. McFarland©2006 5 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. ΔH0f (O2) ΔH 0 f (O3) 4/29/2020 ΔH0f (C, =0 = 142 kJ/mol ΔH0f (C, S.A. McFarland©2006 graphite) = 0 diamond) = 1.90 kJ/mol 6 4/29/2020 S.A. McFarland©2006 7 The standard enthalpy of reaction (ΔH0rxn) is the enthalpy of a reaction carried out at 1 atm. aA + bB 0 ΔHrxn = 0 = ΔHrxn 4/29/2020 cC + dD [cΔH0f (C) + dΔH0f(D) ] - [aΔH0f(A) + bΔH0f (B) ] Σ [nΔH0f (products)] - Σ[ mΔHf0 (reactants)] S.A. McFarland©2006 8 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) H0rxn = S nH0f (products) - S mHf0 (reactants) H0rxn = [ 12H0f (CO2) + 6H0f (H2O)] - [ 2H0f (C6H6)] H0rxn = [ 12(-393.5) + 6(-285.8 )] – [ 2(49.04) ] = -6534 kJ -6534 kJ = - 3267 kJ/mol C6H6 2 mol 4/29/2020 S.A. McFarland©2006 9 Calculate the standard enthalpy of formation of CS2 (l) given that: C(graphite) + O2 (g) CO2 (g) H0rxn = -393.5 kJ S(rhombic) + O2 (g) CS2(l) + 3O2 (g) H0rxn = -296.1 kJ SO2 (g) CO2 (g) + 2SO2 (g) 0 = -1072 kJ Hrxn 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. C(graphite) + O2 (g) 2S(rhombic) + 2O2 (g) + CO2(g) + 2SO2 (g) CO2 (g) H0rxn = -393.5 kJ 2SO2 (g) H0rxn = -296.1x2 kJ CS2 (l) + 3O2 (g) 0 = +1072 kJ Hrxn C(graphite) + 2S(rhombic) CS2 (l) H0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ 4/29/2020 S.A. McFarland©2006 10 The enthalpy of solution (ΔHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. Hsoln = Hsoln - Hcomponents Which substance(s) could be used for melting ice? Which substance(s) could be used for a cold pack? 4/29/2020 S.A. McFarland©2006 11 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. 4/29/2020 S.A. McFarland©2006 13 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 4/29/2020 S.A. McFarland©2006 14 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 λxν=c 4/29/2020 S.A. McFarland©2006 15