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Lecture 4

jan 29 chem lecture 4.docx

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CHEM 1000

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Lecture 4 Jan 29/14 Chemistry 1001 THERMOCHEMISTRY Thermal Energy: KE, random microscopic molecular motions Chemical Energy: PE, relative positions of atoms and molecules, chemicals bonds and intermolecular interactions Internal Energy: U, sum of thermal energy and chemical energy Heat of Reaction: qrxn, heat evolved or absorbed by the reaction mixture Exothermic Reaction: q rxn< 0 Endothermic Reaction: q rxn> 0 A bomb calorimeter is used to carry reactions at constant volume and measure the heat given off by the reaction - see Fig. 7.5 pg 249 (look at this again) C cal: heat capacity of the calorimeter – the steel enclosure + water + stirrer etc. qrxn= -qcal = -C cal T The combustion of 1.010 g sucrose (M = 342.3 g/mol) causes the temperature to rise from 24.92 – 28. 33 C. If the calorimeter constant is 4.90 kJ/K, what is the molar heat of combustion of sucrose? Step 1: find moles of sucrose - We have 1.010/342.3 = 0.00295056 mol sucrose Step 2: the heat of the reaction is - q rxn= -4.90 kJ/K x (28.33 – 24.92) K = - 16.709 kJ - -16.709 kJ / 0.00295056mol = - 5663 kJ/mol Work When a chemical reaction is accompanied by a change of volume of the reaction mixture, it does work (w) by pushing the atmosphere (w > 0) or by yielding to it (w < 0) See fig. 7-7 pg 252 (look at this again – draw pic) Explosions work by quickly releasing a lot of gas (increase n) and/or heat (increases T): Either way, PV (Volume increases) Fig. 7-8 (pressure to volume work – as gas expands it is doing work, f x d = w) - We suddenly remove mass M - The gas pushes the remaining mass M upward by h - W = Fd = ( - Mg) h ( multiply A/A) - A: area of the piston, and P = F/A = Mg / A - w = ( - Mg/A) ( h x A) - w = - Pext V The 2 A’s are distributed in equation NOTE: when V > 0, w <0: the system does work, it loses energy First law of Thermodynamics The energy U of an object can be changed by heat transfer and/ or work U = q + w The internal energy of an isolated system is constant As it is being compressed, a spring evolves 2 J of heat and the work done on the spring is 25 J. by how much did the internal energy of the spring change? a) -27 J b) -23 J c) 23 J v = q + w = -2 + 25 = 23 d) 25 J e) 27 J State function: a function defined at a given time Path function: a function defined for a process (define the actual route taken rather than what exists) that takes places over a time interval Which of these is a path function? a) The distance between my car and York Lanes b) The distance my car travelled in the last 2 days Ex. State functions: T, P, V, U, H … Path functions: q, w Take 0.100mol He at constant T = 298 K. Consider 2 states 1. P 1 2.40 atm V = 1.0189 L 2. P 2 1.20 atm V = 2.0377 L And 2 ways of going from state 1 to state 2 Case 1: The gas expands against P ext= 1.20 atm w 1 - P ext V = - 1.20 atm x 1.0189 L = - 1.22 L atm Case 2: The gas expands (gas doing work) against P ext = 1.80 atm until gas = 1.80 atm AND V = 0.100 x R x 298 / 1.80 = 1.3585 L AND then expands against Pext = 1.20 atm until it reaches state 2 w 2a- 1.80 atm (1.3585 – 1.0189) L = - 0.611 atm w 2b- 1.20 atm (2.0377 – 1.3585) L pv=nrt (insert new pressure) = - 0.815 atm w 2 - 0.611 + (- 0.815) = -1.43 L atm NOTE: that w D2ES NOT EQUAL w 1 More work is done in the 2 step expansion More work would be done in a 3 step expansion And more work is still in a 4 step expansion; etc…. The maximum work is with infinitely many steps each having Pext = Pgas – deltaP This is a reversible expansion (Fig 7-12 pg 258) Consider a reaction Reactants products U
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