Reaction Mechanisms - AP Chemistry
Card 0 of 15
Which of the following is true?
Which of the following is true?
All of the above describe elementary reactions and how they give an overall mechanism.
All of the above describe elementary reactions and how they give an overall mechanism.
Compare your answer with the correct one above
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
and: 
Substitution yields: 
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
Substitution yields:
Compare your answer with the correct one above
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
The reaction can never go faster than its slowest step.
The reaction can never go faster than its slowest step.
Compare your answer with the correct one above
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Since the products are higher in energy than the reactions, the reaction is endothermic.
Since the products are higher in energy than the reactions, the reaction is endothermic.
Compare your answer with the correct one above
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
Compare your answer with the correct one above
Which of the following is true?
Which of the following is true?
All of the above describe elementary reactions and how they give an overall mechanism.
All of the above describe elementary reactions and how they give an overall mechanism.
Compare your answer with the correct one above
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
and:
Substitution yields:
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
Substitution yields:
Compare your answer with the correct one above
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
The reaction can never go faster than its slowest step.
The reaction can never go faster than its slowest step.
Compare your answer with the correct one above
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Since the products are higher in energy than the reactions, the reaction is endothermic.
Since the products are higher in energy than the reactions, the reaction is endothermic.
Compare your answer with the correct one above
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
Compare your answer with the correct one above
Which of the following is true?
Which of the following is true?
All of the above describe elementary reactions and how they give an overall mechanism.
All of the above describe elementary reactions and how they give an overall mechanism.
Compare your answer with the correct one above
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
A possible mechanism for the overall reaction Br2 (g) + 2 NO (g) -> 2 NOBr(g) is
The rate law for the formation of NOBr based on this mechanism is rate = .
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
and:
Substitution yields:
Based on the slowest step the rate law would be: Rate = k2 \[NOBr2\] \[NO\], but one cannot have a rate law in terms of an intermediate (NOBr2).
Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:
Substitution yields:
Compare your answer with the correct one above
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
For the reaction NO2 (g) + CO (g) -> NO (g) + CO2 (g), the reaction was experimentally determined to be Rate = k\[NO2\]2. If the reaction has the following mechanism, what is the rate limiting step, and why?
Step 1: 2 NO2 -> NO3 + NO (slow)
Step 2: NO3 + CO -> NO2 + CO2 (fast)
The reaction can never go faster than its slowest step.
The reaction can never go faster than its slowest step.
Compare your answer with the correct one above
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Based on the figure above, what arrows corresponds to the activation energy of the rate limiting step and the energy of reaction? Is the reaction endo- or exothermic?
Since the products are higher in energy than the reactions, the reaction is endothermic.
Since the products are higher in energy than the reactions, the reaction is endothermic.
Compare your answer with the correct one above
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
Consider the following mechanism:
A + B -> R + C (slow)
A + R -> C (fast)
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
R is the intermediate. It is formed in Step 1 and consumed in Step 2.
Compare your answer with the correct one above