Answer Happy

practice b only:)
7200 = 1 T = Shob- k = A e EXAMPLE 20-9 Applying the Arrhenius Equation Use data from Figure 20-12 to determine the temperature at which t1/2 for the first-order decomposition of N2O5 in CCl4 is 2.00 h. t1 = 1n2 k= in 2 0-693 Analyze 2x60x60 First, find the rate constant k corresponding to a 2.00 h half-life. This can be done by using the half-life for a first-order reaction, In 2 0.693 0.693 k = 9.63 x 10-5s-1 t1/2 2.00 h 7200 s Now, the temperature at which k = 9.63 x 10s can be determined in two ways: graphically and analytically Solve Graphical Method. The temperature at which Ink = In 9.63.1029,248 is marked by the red arrow in Figure 20-12. The value of 1/1 corresponding to Ink = -9.248 is 1/T = 3.28 10-K!, which means that =(x K) = 305 K 3.28 X 10-3K-1 Analytical Method. Take T2 to be the temperature at which k = k2 = 9.63 x 10-55-1. T, is some other tem- perature at which a value of k is known. Suppose we take T1 = 298 K and ki = 3.46 X 10-55-1 referred to in the caption of Figure 20-12. The activation energy is 106 kJ/mol = 1.06 x 10J/mol (the more precise value given in Figure 20-12). Now we can solve equation (20.22) for T2. (For simplicity, we have omit- ted units below, but the temperature is obtained in kelvin.) k2 Ea 1 RT2 9.63 x 10-5 1.06 X 105 1 1 In 3.46 X 10-5 8.3145 T2 298 1 1.27 X 104 1.024 = -1.27 x 104 0.00336 = 42.7 - T2 T2 1.27 X 104 = 42.7 - 1.024 = 41.7 T2 (1.27 x 104) T2 = = 305 K a point In ki =) 1046 41.7 Assess Both methods agree extremely well in this case. Depending on the circumstance one method may be preferred over the other. u3 PRACTICE EXAMPLE A: What is the half-life of the first-order decomposition of N2O5 at 75.0°C? Use data from Example 20-9. In Kr_ - t/R (21/11) PRACTICE EXAMPLE B: At what temperature will it take 1.50 h foot two-thirds of a sample of N2O; in Cal, to decompose in Example 20-9? Inrt/2