test2s-1 - (c The first and second-order graphs are shown...

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Chemistry 2740 Spring 2010 Test 2 Solutions 1. (a) The cell emf (b) The emf for a cell under standard conditions (c) The number of electrons exchanged in the redox process for the reaction as written 2. We need the activity of the hydrogen ion. HCl is a strong acid, so we know the concentration of hydrogen ions in this solution. We just need the activity coefficient. I c = 1 2 { (+1) 2 [H + ] + ( - 1) 2 [Cl ] } = 1 2 { 0 . 0058 + 0 . 0058 mol / L } = 0 . 0058 mol / L . ln γ H + = - A (+1) 2 ( εT ) 3 / 2 I c = - 1 . 107 × 10 10 [ (6 . 957 × 10 10 C 2 N 1 m 2 )(298 . 15 K) ] 3 / 2 0 . 0058 mol / L = - 0 . 089 . γ H + = e 0 . 089 = 0 . 915 . a H + = γ H + [H + ] /c = 0 . 915(0 . 0058) = 0 . 0053 . p H = - log 10 a H + = - log 10 (0 . 0053) = 2 . 28 . 3. (a) Reduction (b) According to the law of mass action, the first reaction should have a second-order rate law, while the second reaction should have a first-order rate law. In the latter case, a first-order rate law arises because the solvent is in great excess. The reaction thus has no real effect on its concentration.
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Unformatted text preview: (c) The first- and second-order graphs are shown in figures 1 and 2, respectively. It is clear from figure 2 that the data do not fit a second-order rate law. We can therefore eliminate mechanism (i) as a possibility. Opinions may differ on the linearity of the data seen in figure 1. Personally, I think that the data display reasonable linearity. I would therefore conclude that (ii) is the likely mechanism. The slope of this graph is-. 056 s − 1 . The rate constant is therefore 0 . 056 s − 1 . 1-3-2.5-2-1.5-1-0.5 0.5 1 1.5 10 20 30 40 50 60 70 80 90 ln I t /s Figure 1: First-order plot for the data of question 3c. 2 4 6 8 10 12 10 20 30 40 50 60 70 80 90-1 t /s Figure 2: Second-order plot for the data of question 3c. 2...
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