Kinetics - College Chemistry
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Which of the following is true regarding an enzyme-catalyzed reaction?
Which of the following is true regarding an enzyme-catalyzed reaction?
For this question, we need to determine a true statement with regard to reactions that are catalyzed by enzymes.
To answer this, it's important to distinguish between the thermodynamics of a reaction and the kinetics. When adding an enzyme, the activation energy for the reaction is lowered. What this means is that it becomes easier for the reactants to achieve the high-energy transition state on their way to becoming product. Thus, enzymes affect the kinetics of a reaction by increasing both the forward rate and the reverse rate.
But if we look at the thermodynamics of an enzyme-catalyzed reaction, there is no change. In other words, the difference in free energy between the reactants and the products remains the same, regardless of the presence of enzyme. Thus, an enzyme will not cause a reaction to change its equilibrium position; it cannot shift a reaction towards the left or towards the right.
Because the thermodynamics of the reaction remain unchanged, the reaction will not become more exergonic, nor will it become more endergonic. Furthermore, the equilibrium of the reaction will not shift.
For this question, we need to determine a true statement with regard to reactions that are catalyzed by enzymes.
To answer this, it's important to distinguish between the thermodynamics of a reaction and the kinetics. When adding an enzyme, the activation energy for the reaction is lowered. What this means is that it becomes easier for the reactants to achieve the high-energy transition state on their way to becoming product. Thus, enzymes affect the kinetics of a reaction by increasing both the forward rate and the reverse rate.
But if we look at the thermodynamics of an enzyme-catalyzed reaction, there is no change. In other words, the difference in free energy between the reactants and the products remains the same, regardless of the presence of enzyme. Thus, an enzyme will not cause a reaction to change its equilibrium position; it cannot shift a reaction towards the left or towards the right.
Because the thermodynamics of the reaction remain unchanged, the reaction will not become more exergonic, nor will it become more endergonic. Furthermore, the equilibrium of the reaction will not shift.
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Which of the following statements is true with respect to reactions involving enzymes?
Which of the following statements is true with respect to reactions involving enzymes?
For this question, we're asked to select a true statement regarding reactions that are influenced by enzymes.
When thinking about how enzymes affect reactions, its important to draw a distinction between kinetics and thermodynamics. Enzymes increase a reaction's rate, but they don't change the equilibrium of a reaction. In other words, they make reactions go faster, but they don't make reactions go.
Since enzymes don't affect equilibrium, we can rule out two of the answer choices. Moreover, we know that enzymes increase the reaction rate by lower the activation energy. Consequently, this raises the rate constant for the reaction.
For this question, we're asked to select a true statement regarding reactions that are influenced by enzymes.
When thinking about how enzymes affect reactions, its important to draw a distinction between kinetics and thermodynamics. Enzymes increase a reaction's rate, but they don't change the equilibrium of a reaction. In other words, they make reactions go faster, but they don't make reactions go.
Since enzymes don't affect equilibrium, we can rule out two of the answer choices. Moreover, we know that enzymes increase the reaction rate by lower the activation energy. Consequently, this raises the rate constant for the reaction.
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Which of the following techniques will not increase the rate of a reaction?
Which of the following techniques will not increase the rate of a reaction?
Increasing the amount of reactant increases the rate of reaction only if the concentration of each reactant increases and if the reactants are dissolved in a solution. This is not the case for all reactions. Heating the reaction and introducing a catalyst enable the reactants to reach the activation energy needed to proceed into the reaction. Increasing the surface area between the reactants enables more of each reactant to interact with one another.
Increasing the amount of reactant increases the rate of reaction only if the concentration of each reactant increases and if the reactants are dissolved in a solution. This is not the case for all reactions. Heating the reaction and introducing a catalyst enable the reactants to reach the activation energy needed to proceed into the reaction. Increasing the surface area between the reactants enables more of each reactant to interact with one another.
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If a reaction is zero order with respect to 
, which of the following quantities will produce a straight line when graphed against 
(time)?
If a reaction is zero order with respect to
The correct answer is 
because the differential rate law for a zero order reaction is 
. Therefore, the plot has a linear relationship to the concentration of hydrogen.
The correct answer is
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If a reaction is first order with respect to 
, which of the following quantities will produce a straight line when graphed against 
(time)?
If a reaction is first order with respect to
Because the differential rate law for a first order reaction is 
the graph will have a linear relationship to 
.
Because the differential rate law for a first order reaction is
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If a reaction is second order with respect to 
, which of the following quantities will produce a straight line when graphed against 
(time)?
If a reaction is second order with respect to
Because the differential rate law for a second order equation is 
the graph will have a linear relationship to 
.
Because the differential rate law for a second order equation is
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Using the reaction and experimental data given, determine the rate equation for this reaction.
Using the reaction and experimental data given, determine the rate equation for this reaction.
In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to 
. When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to 
. Then, we plug in all of this information into the rate law, and we get the correct answer of 
.
In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to
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Given the equation and experimental data above determine the value of the rate constant 
for this reaction.
Given the equation and experimental data above determine the value of the rate constant
In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to 
. When comparing experiments 1 and 3, we see that the concentration of B was doubled, and this resulted in a doubling of the rate. Therefore, the reaction is first order with respect to 
. Then, we plug in all of this information into the rate law, and we get the correct answer of 
.
Finally, we take the values from one of the rows of the experimental data. It does not matter which one, they will all result in the same answer. In this case, I chose experiment 1.
Because the values from the table all only have one significant figure, our answer can also only have one significant figure.
In order to solve this problem we must compare the initial concentrations of our two reactants between equations with how they effect the rate of the reaction. When comparing experiment 1 and 2, we see that the concentration of A was doubled. This resulted in a quadrupling of the rate. Therefore the reaction is second order with respect to
Finally, we take the values from one of the rows of the experimental data. It does not matter which one, they will all result in the same answer. In this case, I chose experiment 1.
Because the values from the table all only have one significant figure, our answer can also only have one significant figure.
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What are the correct units for the rate constant 
of a zero order reaction?
What are the correct units for the rate constant
Because the rate of a chemical reaction is always in 
. The 
value must always cancel with other units to produce these units. In a zero order reaction there is nothing to cancel, so the units are simply 
.
Because the rate of a chemical reaction is always in
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What are the units for the rate constant 
for a first order reaction?
What are the units for the rate constant
Because the rate of a chemical reaction is always in 
. The k value must always cancel with other units to produce these units. In a first order reaction the rate equation is
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match, 
must simply be 
.
Because the rate of a chemical reaction is always in
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match,
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What are the units for the rate constant 
for a second order reaction?
What are the units for the rate constant
Because the rate of a chemical reaction is always in 
. The k value must always cancel with other units to produce these units. In a second order reaction the rate equation is
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match, 
must be have the units 
.
Because the rate of a chemical reaction is always in
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match,
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What are the units for the rate constant 
of a third order reaction?
What are the units for the rate constant
Because the rate of a chemical reaction is always in 
. The 
value must always cancel with other units to produce these units. In a third order reaction the rate equation is
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match, 
must be have the units 
.
Because the rate of a chemical reaction is always in
If we replace the variables above with their units, the equation will look like this:
Therefore, in order to match,
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The reaction
Follows the mechanism:
(slow)
(fast)
Determine the rate law for this reaction.
The reaction
Follows the mechanism:
Determine the rate law for this reaction.
Remember, the rate law is always determined by the rate-determining step. This is ALWAYS the "slow" step in the reaction. In this case, the slow step is step (i).
The rate for this step alone is
And because both of the original reactants are in this rate law, it is also the rate law for the overall reaction.
Remember, the rate law is always determined by the rate-determining step. This is ALWAYS the "slow" step in the reaction. In this case, the slow step is step (i).
The rate for this step alone is
And because both of the original reactants are in this rate law, it is also the rate law for the overall reaction.
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If the reaction:
Follows the mechanism:
(Slow)
(Fast)
What is the rate order with respect to 
?
If the reaction:
Follows the mechanism:
What is the rate order with respect to
Because B is in the rate determining step (the slow step) then we know that it is not zero order with respect to the concentration of B.
However, because there is only one molecule of B reacting in the rate determining step, we know that this reaction is first order with respect to the concentration of B.
Because B is in the rate determining step (the slow step) then we know that it is not zero order with respect to the concentration of B.
However, because there is only one molecule of B reacting in the rate determining step, we know that this reaction is first order with respect to the concentration of B.
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If the reaction:
Follows the mechanism:
(Slow)
(Fast)
What is the rate order with respect to 
?
If the reaction:
Follows the mechanism:
What is the rate order with respect to
The rate law is determined by the slowest step in a reaction mechanism. Because B is not a reactant in the slow step, then it will not appear in the reaction mechanism.
Therefore, the reaction is zeroth order in relation to the concentration of B.
The rate law is determined by the slowest step in a reaction mechanism. Because B is not a reactant in the slow step, then it will not appear in the reaction mechanism.
Therefore, the reaction is zeroth order in relation to the concentration of B.
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