University:
Amherst College
Course:
CHEM 161 | Chemical Principles
Academic year:

2009
Views:

68

Pages:

6
Author:

ice0ver13toy

Chemistry 12 Assignment 6 - Solutions Fall 2009 Problems: Chapter 10: #71, 72, 73, 74, 90, 92, 109 Chapter 6: #17, 20, 63, 78 FREE ENERGY: Pressure Dependence and Equilibrium At 25.0°C: ΔG° = ΔH° - TΔS° = -58.03 x 10⁹ J/mol - (298.2 K)(-176.6 J/Kmol) = -5.37 x 10³ J/mol ln K = -ΔG°/RT = -(-5.37 x 10³ J/mol) / (8.3145 J/Kmol)(298.2 K) = 2.166; K = e^2.166 = 8.72 At 100.0°C: ΔG° = -58.03 x 10⁹ J/mol - (373.2 K)(-176.6 J/Kmol) = 7.88 x 10³ J/mol ln K = -(7.88 x 10³ J/mol) / (8.3145 J/Kmol)(373.2 K) = -2.540; K = e^-2.540 = 0.0789 The equation ln K = -ΔH°/R(1/T) + ΔS°/R is in the form of a straight line equation (y = mx + b). A graph of ln K vs. 1/T will yield a straight line with slope m = -ΔH°/R and y-intercept b = ΔS°/R. From the plot: slope = -1.3 x 10⁴ K y-intercept = 40. ΔS°/R = 40, ΔS° = 40 x 8.3145 J/Kmol = 330 J/Kmol As seen here, when ΔH° is positive, the slope of the ln K vs. 1/T plot is negative. When ΔH° is negative, as in an exothermic process, the slope of the ln K vs. 1/T plot will be positive (slope = -ΔH°/R). A graph of ln K vs. 1/T will yield a straight line with slope equal to -ΔH°/R and y-intercept equal to ΔS°/R (see Exercise 10.72). Temp (°C) T(K) 1000/T (K⁻¹) Kw ln Kw 0 273 3.66 1.14 x 10⁻¹⁴ -34.408 25 298 3.36 1.00 x 10⁻¹⁴ -32.236 35 308 3.25 2.09 x 10⁻¹⁴ -31.499 40 313 3.19 2.92 x 10⁻¹⁴ -31.165 50 323 3.10 5.47 x 10⁻¹⁴ -30.537 Graph A plot of ln Kw vs. 1000/T (K⁻¹) Calculations The straight line equation (from a calculator) is: ln Kw = -6.91 x 10³ (1/T) - 9.09 Slope = -6.91 x 10³ K = -ΔH°/R ΔH° = (-6.91 x 10³ K) * (8.3145 J K⁻¹ mol⁻¹) = 57.5 kJ/mol y-intercept = -9.09 = ΔS°/R ΔS° = -9.09 * 8.3145 J K⁻¹ mol⁻¹ = -75.6 J K⁻¹ mol⁻¹ Part b: From part a, ΔH° = 57.5 kJ/mol and ΔS° = -75.6 J K⁻¹ mol⁻¹. Assuming ΔH° and ΔS° are temperature independent: ΔG° = ΔH° - TΔS° = 57,500 J/mol - 647 K (-75.6 J K⁻¹ mol⁻¹) = 106,400 J/mol = 106.4 kJ/mol Subtracting the first expression from the second: ln K₂ - ln K₁ = (ΔH°/R)(1/T₁ - 1/T₂) ln(K₂/K₁) = (ΔH°/R)(1/T₁ - 1/T₂) For K = 8.84 at T = 25 °C: ln(3.25 x 10⁻²/8.84) = (ΔH°/8.3145 J K⁻¹ mol⁻¹)(1/298 K - 1/348 K) -5.61 = (5.8 x 10⁴ J/mol)(ΔH°) ΔH° = -9.7 x 10⁴ J/mol For K = 8.84 at T = 25 °C: ΔS° = (ΔH° - ΔG°) / T ΔS° = (-9.7 x 10⁴ J/mol) / (8.3145 J K⁻¹ mol⁻¹)(298 K) ΔS° = -310 J K⁻¹ mol⁻¹ We get the same value for ΔS° using K = 3.25 x 10⁻² and T = 348 K data. ΔG° = -RT ln K. When K = 1.00, then ΔG° = 0 since ln 1.00 = 0. ΔG° = 0 = ΔH° - TΔS°. Assuming ΔH° and ΔS° do not depend on temperature: ΔH° = TΔS° T = ΔH°/ΔS° = (-9.

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