Chapter 13: Problem 63
What is the rule of thumb for how reaction rate changes as the temperature changes?
Chapter 13: Problem 63
What is the rule of thumb for how reaction rate changes as the temperature changes?
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Get started for freeThe slowest step in a reaction mechanism is called the _______-_______ step.
What is meant by the following terms: (a) Substrate (b) Active site (c) Lock-and-key mechanism
The experimental rate law for the reaction \(\mathrm{A}+\mathrm{A} \rightarrow \mathrm{A}_{2}\) is Rate \(=k[\mathrm{~A}][\mathrm{BC}]\) Two mechanisms have been proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB}\) (slow) Step \(2: \mathrm{AB}+\mathrm{A} \rightarrow \mathrm{A}_{2}+\mathrm{B}\) (fast) and Step \(1: \mathrm{A}+\mathrm{BC} \rightarrow \mathrm{AB}+\mathrm{C}\) (slow) Step \(2: \mathrm{A}+\mathrm{AB} \rightarrow \mathrm{B}+\mathrm{A}_{2}\) (fast) Step \(3: \mathrm{B}+\mathrm{C} \rightarrow \mathrm{BC}\) (fast) (a) Show that each mechanism results in the correct overall reaction. (b) Which mechanism is consistent with the rate law? (c) Why does \(\mathrm{BC}\) appear in the rate law but not in the overall reaction?
For a particular reaction, the absorbed energy is \(800 \mathrm{~kJ}\) to break old bonds, and \(\Delta E_{\mathrm{rxn}}\) is equal to \(-800 \mathrm{~kJ}\). How much energy is released into the surroundings as the product bonds are formed?
Why can we ignore other steps and use only the rate-determining step in a mechanism to write the predicted rate law?
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