122ch15c - 41 15.68) Kp = Kc (RT)n Kc = Kp (RT)&)n =...

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41 15.68) K p = K c (RT) ) n K c = K p (RT) ) n = (4.3 3 2) {(0.0821 L C atm/mol C K)(1100 K)} ( 1) = 0.047 9 68 = 0.0480
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42 BONUS PROBLEM: (especially for the engineers):
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43 15.71) Calculate the initial conc. of CO 2 and H 2 and use equil. (ICE) table to find equil. conc. [CO 2 ] = [H 2 ] = 1.50 mol/0.750 L = 2.00 M CO 2 (g) + H 2 (g) W CO (g) + H 2 O (g) I nitial 2.00 2.00 0 0 C hange ! x ! x + x + x ---------------------------------------------------------------------- E quil (2.00 ! x) (2.00 ! x) x x [CO] [H 2 O] K c = ---------------- = 0.802 [CO 2 ] [H 2 ] (x) 2 (x) 2 K c = --------------------------- = --------------- = 0.802 (2.00 ! x) (2.00 ! x) (2.00 ! x) 2 This eqn is a perfect square - we can take the square root of both sides. (x) --- ------------ = (0.802) 1/2 = 0.89 5 54 (2.00 ! x) 0.89 5 54 (2.00 ! x) = x 1.7 9 10 ! (0.89 5 54)x = x 1.7 9 10 = (1.89 5 54)x x = 0.94 4 8 = 0.945 M [CO 2 ] = (2.00 ! x) = (2.00 ! 0.94 4 8) = 1.0 5 51 = 1.06 M [H 2 ] = (2.00 ! x) = (2.00 ! 0.94 4 8) = 1.0 5 51 = 1.06 M [CO] = x = 0.94 4 8 = 0.945 M [H 2 O] = x = 0.94 4 8 = 0.945 M Check: K c = (0.94 4 8) 2 /(1.0 5 51) 2 = 0.802
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44 15.73)
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45 15.74)
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46 15.76) Mathematical LeChatelier’s Principle Application
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47 15.78)
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48 15.79) The patent claim is FALSE. A catalyst does NOT alter the position of equilibrium, ONLY the rate of reaching equilibrium. 15.82)
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49 15.83)
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50 15.84) Both the forward and reverse reactions in the equilibrium A º B are elementary (single-step) reactions. Assume the only effect a catalyst has on the reaction is to
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This note was uploaded on 06/11/2011 for the course CHEM 122 taught by Professor Zellmer during the Spring '07 term at Ohio State.

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122ch15c - 41 15.68) Kp = Kc (RT)n Kc = Kp (RT)&)n =...

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