Exam 2 key

Exam 2 key - Chem 10 Exam 2 version A Useful information: R...

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Chem 10 – Exam 2 version A Useful information: R = 0.08206 liter atm/ mol K = 8.3145 J/mol K 1 atm = 101325 Pa = 760 torr, F=95485 coul/mol, RT/F (T=298 K) = 0.591 V Reduction potentials: Ni 2+ (aq) + 2e Ni(s) 0.25 Pb 2+ (aq) + 2e Pb(s) 0.13 I 2 (s) + 2e 2I (aq) +0.54 Br 2 (aq) + 2e 2Br (aq) +1.09 Cl 2 (g) + 2e 2Cl (aq) +1.36 Cr 2 O 7 2 (aq) + 14H + + 6e 2Cr 3+ (aq) + 7H 2 O +1.33 Consider the reaction N 2 O 4 (g) 2NO 2 (g) At 25 ºC, the values of Δ Hº and Δ Sº are 58.3 kJ/mol and 176.6 J mol -1 K -1 , respectively. 1. Do you expect higher temperature to favor reactants or products? a) reactants b) products c) no effect d) not enough information to determine 2. Assuming Δ Hº and Δ Sº are temperature independent, calculate the values of K eq at 150 ºC. lnK eq = - Δ H/RT + Δ S/R = 58300 J/(8.314J mol -1 K -1 x 423 K) + 199 J mol -1 K -1 / 8.314 J mol -1 K -1 = 4.66 or K eq = 106 3.
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This note was uploaded on 07/02/2010 for the course CHEM 10 taught by Professor G during the Spring '10 term at UCM.

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Exam 2 key - Chem 10 Exam 2 version A Useful information: R...

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