Lecture 39 - Topics for the day Administrative stuff...

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Topics for the day Administrative stuff Homework problems Constant volume calorimetry Spontaneity Entropy Entropy as probability Administrative stuff Quest site had upload problems yesterday As announced, HW5 due date delayed till 5 pm today You’ll be notiFed when HW6 (last one) is posted ALEKS Obj 6 is over, Obj 7 due ±riday 5/7 midnight I’m removing some of the Obj 7 topics to shorten it. Decision who gets the option of skipping the Fnal Can’t be made until after 5/7 after last ALEKS is done. But before ALEKS pie is opened Where were we? Constant pressure calorimetry measures Δ H for reactions that involve no gases. Thermal energy released by the reaction is transferred to the solution and the calorimeter: Δ H = q P The Internal energy E of a substance is the total of all the energy (both kinetic and potential) contained within that substance. Work w = –P Δ V is one way a system can exchange energy with the surroundings. In chemistry, work is only done during reactions involving gases, when a change of volume occurs. Expansion is work done by the system (–w), compression is work done on the system (+w). Work is NOT a state function. It is dependent on pathway. The two ways a system can exchange energy are by heat and work. Δ E = q + w = q – P Δ V Now we can deFne enthalpy: Δ H = Δ E + Δ n gas RT 1 2 3
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A. Δ H is larger than Δ E B. Δ E is larger than Δ H C. Δ H is equal to Δ E iClicker Time Consider this reaction: C(s) + O 2 (g) CO 2 (g) Which statement is true? Δ n gas = n f – n i = 1 – 1 = 0 Δ H = Δ E + Δ n gas RT A balloon is fully inFated by heating the air inside it. The balloon volume changes from 4.00 × 10 6 L to 4.50 × 10 6 L by the addition of 1.3 × 10 8 J of energy as heat. Assuming the balloon expands against a constant atmospheric pressure of 1.0 atm, calculate the change in internal energy for the process. Given 1 L atm = 101.325 J Calculation time Δ E = q + w Heat added to system (i.e., positive): q = +1.3 × 10 8 J w = – P Δ V = – P (V f – V i ) = –1 atm × (4.50 × 10 6 L – 4.00 × 10 6 L) = – 0.50 × 10 6 L atm Sign check – expansion is work done BY system, (i.e., energy lost from the system), so it should be negative. To add together q and w, they need to be in the same units. w = – 0.50 × 10 6 L atm × 101.325 J L –1 atm –1 = –5.1 × 10 7 J Δ E = q + w = 1.3 × 10 8 J + –5.1 × 10 7 J = 7.9 × 10 7 J More energy is added to the system by heating, than is used up by the gas doing work expanding. There is a net increase in the internal energy of the system. Δ E is positive.
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This note was uploaded on 08/27/2011 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas.

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Lecture 39 - Topics for the day Administrative stuff...

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