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Unformatted text preview: time (x 10 3 s) Name: 2 Part II: Calculating the Activation Energy. II.1 (18 points) The same reaction from Part I was monitored at different temperatures, and rate constants were determined as done in Part I. These data are presented in the following table. N 2 O 5 (g) → 2 NO 2 (g) + ½ O 2 (g) T (K) k (s-1 ) 318 5.0 x 10-4 308 1.4 x 10-4 298 3.5 x 10-5 Using the data from the table above, the value of k you calculated in Part I, and the graph below, calculate the activation energy (E a ) of the reaction. II.2 (10 points) What is the half-life of the reaction at 298 K? Show derivation of the equation starting with the integrated rate law....
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- Spring '08
- Chemistry, Chemical reaction, Cornell University, Rate equation, Cornell University Dr.