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40 Chapter 2 STRUCTURE AND REACTIVITY 50. (a) From 44(a), we have = -1.30 kcal mol⁻¹. T = 298 K and = +1.4 cal mol⁻¹ = +1.4 X 10⁻³ kcal deg⁻¹ mol⁻¹. So = TAS° needs to be rearranged to solve for = + = 1.30 + X 10⁻³) = 1.30 + 0.42. = -0.88 kcal mol⁻¹. This agrees very nicely with the = -0.9 kcal mol⁻¹ calculated in Problem 44(b),(c) on page 36 from the number of gauche interactions in the 0° conformation relative to the 120° conformation. (b) Don't forget to change °C to K by adding 273°! (1) (-250°C) = - = -0.88 (23K)(1.4 X 10⁻³) = -0.91 kcal mol⁻¹ (2) (-100°C) = = -0.88 - (1.73K)(1.4 X 10⁻³) = 1.12 kcal mol⁻¹ (3) (500°C) = = -0.88 (773K)(1.4 X 10⁻³) = 1.96 kcal mol⁻¹ (c) Use = RT In K = -2.303 RT log K. This rearranges to 2.303RT = log K, or K = = (1) At T = -250°C = 23 K, = -0.91 kcal = -910 cal mol = 2.303RT -910 = 8.65 = log K, so K = 4.5 X 10⁸. 2.303(1.986)(23) (2) At T = 100°C = 173 K, = -1.12 kcal = 1120 cal mol⁻¹; = 2.303RT - 1120 = 1.42 = log K, so K = 26. 2.303(1.986)(173) (3) At T = 500°C = 773 K, = 1.96 kcal = 1960 cal mol⁻¹; = -1960 - 2.303RT = 0.55 = log K, so K = 3.5. 2.303(1.986)(773) We can summarize the results of Problems 50 and 44 in a little table: T(K) K 23 -0.91 4.5 X 10⁸ 173 -1.12 26 298 -1.30 9 773 -1.96 3.5 These data illustrate two points. The most obvious is the huge effect of temperature on K. At 23 K (which is very cold), only two 2-methylbutane molecules out of every billion are in the higher energy (0°) confor- mation! There is very little thermal energy around to cause bond rotation to occur. In contrast, at higher tem- peratures K drops as the increased thermal energy allows more and more molecules to be in less stable conformations. Note also that the value does cause G° to vary with temperature, too, but the effect is small, since is small.