# Assignment 3 2010-11 Fall Answers - CHEM 241. Assignment 3:...

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CHEM 241. Assignment 3: Thermochemistry ANSWERS Hand in: Thursday October 21 in class. No late submissions accepted, no e-mail submissions. Answer questions 1 -3 in the spaces below, question 4 on one separate sheet. Show your work! 1. (7 points) When 0.985 g of benzoic acid C 7 H 6 O 2 (s) was combusted in a bomb calorimeter (closed and constant volume), the temperature rose from 21.34 to 27.31 o C. The calorimeter constant (i.e., the heat capacity of the complete calorimeter and content) is 4.351x10 3 J.K −1 . a) (4 points) Calculate the standard molar energy of combustion, ΔU o m,combustion of benzoic acid. Answer. The bomb calorimeter is an adiabatic, constant volume system, so q = q calorimeter + q V,combustion = 0, therefore q V,combustion = ΔU combustion = − q calorimeter q calorimeter = CΔT = 4.351x10 3 (27.31−21.34) = 2.598x10 4 J; q V,combustion = −2.598x10 4 J So for 0.985 g benzoic acid ΔU = −2,598x10 4 J For 1 mol: ΔU m, combustion = (122.0/0.985)x(−2,598x10 4 ) = −3.217.10 6 J/mol or −3.217x10 3 kJ/mol There might be some question about significant figures/error limits. The 0.985 has only 3 sign. Figs., but we can also say it is 0.985 ±0.001 g or 0.985±0.1% . More importantly, the ΔT = 5.97±0.2, a larger uncertainty of 0.3%, therefore we should report the answer only with 3 significant figures as –3.22x10 6 J/mol ±0.3% or (3.22±0.01)x10 3 kJ Answer: ΔU o m,combustion = −3.22x10 6 J/mol b) (1 point) Calculate the standard molar enthalpy of combustion, ΔH o combustion of benzoic acid. Hint: first balance the combustion reaction. Remember that ΔpV = ΔnRT ≈ RTΔn gas . Answer . The balanced reaction is: C 7 H 6 O 2 (s) + 7.5O 2 (g) → 7CO 2 (g) + 3H 2 O(l) We find Δn gas = –0.5 mol. This is for 1 mol benzoic acid, so we use ΔU m and calculate ΔH m . For the moment we keep the three decimal places in ΔU, then: ΔH m = ΔU m + pΔV = ΔU + RTΔn gas = –3.217x10 6 + 8.314x297.4x(–0.5) = –3.217x10 6 –1.236x10 3 = –3.218x10 6 J/mol. Note how infinitesimally small the RTΔn correction is, it is well within the error limit of the ±0.01. This is very common in these combustion experiments: if the data have error limits of more than 0.1%, there actually is no need for the rtΔn correction, and for all practical purposes ΔU = ΔH! Answer:

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## This note was uploaded on 10/27/2010 for the course ENGINEERIN 200500567 taught by Professor Ahmed during the Spring '10 term at Qatar University.

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Assignment 3 2010-11 Fall Answers - CHEM 241. Assignment 3:...

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