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

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Chapter1: Chemical Foundations
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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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