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Consider two experiments, with identical conditions except for one experiment contained a catalyst. Mark all that apply to this situation. Experiment 1: E, - 19 kJ = Experiment 2: E, 12 kJ Experiment 1 used the catalyst Experiment 2 used the catalyst Experiment 1 reaction was faster Experiment 2 reaction was faster Experiment 1 has a higher activation energy barrier Experiment 2 has a higher activation energy barrier

Consider two experiments, with identical conditions except for one experiment contained a catalyst. Mark all that apply to this situation. Experiment 1: E, - 19 kJ = Experiment 2: E, 12 kJ Experiment 1 used the catalyst Experiment 2 used the catalyst Experiment 1 reaction was faster Experiment 2 reaction was faster Experiment 1 has a higher activation energy barrier Experiment 2 has a higher activation energy barrier

Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter11: Rate Of Reaction
Section: Chapter Questions
Problem 71QAP: For a certain reaction, Ea is 135 kJ and H=45 kJ. In the presence of a catalyst, the activation...
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Consider two experiments, with identical conditions except for one experiment contained a catalyst. Mark all that apply to this situation. Experiment 1: E = 19 kJ Experiment 2: E,- 12 kJ Experiment 1 used the catalyst Experiment 2 used the catalyst Experiment 1 reaction was faster Experiment 2 reaction was faster Experiment 1 has a higher activation energy barrier Experiment 2 has a higher activation energy barrier
Transcribed Image Text:Consider two experiments, with identical conditions except for one experiment contained a catalyst. Mark all that apply to this situation. Experiment 1: E = 19 kJ Experiment 2: E,- 12 kJ Experiment 1 used the catalyst Experiment 2 used the catalyst Experiment 1 reaction was faster Experiment 2 reaction was faster Experiment 1 has a higher activation energy barrier Experiment 2 has a higher activation energy barrier
For a plot of In(k) vs 1/T, how is activation energy related to slope? O E, m O E₂ --m O E₂ = -m/R E = -m R For a forward reaction, E, - 5 kJ and AH = -20. kJ. Mark all that apply to this situation. (Hint: sketch a PE plot so you can see what's going on). The reaction is endothermic The reaction is exothermic The forward rate is probably faster than reverse rate The reverse rate is probably faster than the forward rate Activation energy (reverse reaction) = -5 kJ Activation energy (reverse reaction) - 20 kJ Activation energy (reverse reaction) - 25 kJ O Enthalpy (reverse reaction) = -5 kJ Activation energy (reverse reaction) - 20 kJ D Activation energy (reverse reaction) - 25 kJ
Transcribed Image Text:For a plot of In(k) vs 1/T, how is activation energy related to slope? O E, m O E₂ --m O E₂ = -m/R E = -m R For a forward reaction, E, - 5 kJ and AH = -20. kJ. Mark all that apply to this situation. (Hint: sketch a PE plot so you can see what's going on). The reaction is endothermic The reaction is exothermic The forward rate is probably faster than reverse rate The reverse rate is probably faster than the forward rate Activation energy (reverse reaction) = -5 kJ Activation energy (reverse reaction) - 20 kJ Activation energy (reverse reaction) - 25 kJ O Enthalpy (reverse reaction) = -5 kJ Activation energy (reverse reaction) - 20 kJ D Activation energy (reverse reaction) - 25 kJ
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