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Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s
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Majid Nabavi
Determine the order of the reaction, and the value of the rate constant. Predict the concentration of AB at 25s
Whenever you are given concentrations vs. time, think "integrated rate law." There are three integrated rate laws that your instructor will either give you or expect you to know:
(1) for a zero order reaction: $[A]_t = -kt + [A]_0$
(2) for a 1st order reaction: $ln[A]_t = -kt + ln[A]_0$
(3) for a 2nd order reaction: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$
An important thing to note is that each of these equations is in the same pattern as the equation for a line, viz., $y=mx+b$. For example, number 2 above (for the 1st order equation):
Why is this important? Because you can plot the data as expected by each of these equations and the one that plots as a straight line tells you the order of the reaction.
So, use your calculator or spreadsheet and enter the data. Then add two columns. One column computes the natural log of the given concentrations. The second column computes the reciprocal of the given concentrations.
Now create three separate plots. Plot "Time" on the horizontal axis of all three plots. Then, on the vertical axis of each separate chart, plot the given concentration in chart1, the natural log of concentration in chart2 and the reciprocal of concentration in chart3. You will find that chart3 is the graph of a straight line. Therefore, this is a second order reaction.
Your spreadsheet or calculator can calculate a linear regression line. The value of $k$ is the slope of that linear regression line. How do we know that? Because $k$ is the slope for each of the integrated rate law equations. In this case, the integrated rate law for this second order reaction is
as shown above. (The $kt$ term is the $mx$ term in the equation for a line and $m$ is the slope.)
Finally, you can calculate the concentration of AB at 25 seconds using the equation for the regression line, which in this case is $y=0.0225x + 1.06$. Since we calculated the regression line for $\frac{1}{[AB]}$ we need to reverse that, so we want $[AB]_{25s} = \frac{1}{y}$:
$[AB]=\frac{1}{0.0225 \cdot 25 + 1.06}$ (the regression calculated the reciprocal so we need to reverse that)
Whenever you are given concentrations vs. time, think "integrated rate law." There are three integrated rate laws that your instructor will either give you or expect you to know:
(1) for a zero order reaction: $[A]_t = -kt + [A]_0$
(2) for a 1st order reaction: $ln[A]_t = -kt + ln[A]_0$
(3) for a 2nd order reaction: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$
An important thing to note is that each of these equations is in the same pattern as the equation for a line, viz., $y=mx+b$. For example, number 2 above (for the 1st order equation):
Why is this important? Because you can plot the data as expected by each of these equations and the one that plots as a straight line tells you the order of the reaction.
So, use your calculator or spreadsheet and enter the data. Then add two columns. One column computes the natural log of the given concentrations. The second column computes the reciprocal of the given concentrations.
Now create three separate plots. Plot "Time" on the horizontal axis of all three plots. Then, on the vertical axis of each separate chart, plot the given concentration in chart1, the natural log of concentration in chart2 and the reciprocal of concentration in chart3. You will find that chart3 is the graph of a straight line. Therefore, this is a second order reaction.
Your spreadsheet or calculator can calculate a linear regression line. The value of $k$ is the slope of that linear regression line. How do we know that? Because $k$ is the slope for each of the integrated rate law equations. In this case, the integrated rate law for this second order reaction is
as shown above. (The $kt$ term is the $mx$ term in the equation for a line and $m$ is the slope.)
Finally, you can calculate the concentration of AB at 25 seconds using the equation for the regression line, which in this case is $y=0.0225x + 1.06$. Since we calculated the regression line for $\frac{1}{[AB]}$ we need to reverse that, so we want $[AB]_{25s} = \frac{1}{y}$:
$[AB]=\frac{1}{0.0225 \cdot 25 + 1.06}$ (the regression calculated the reciprocal so we need to reverse that)
If n is the order of this reaction, then the rate equation is given by
$$-\frac{d[AB]}{dt}=k[AB]^n$$
If we take the log of both sides of this equation, we obtain:
$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$
This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$. So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk. So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph? Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval. If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:
$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$
The average concentration over this same time interval would be:
$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$
So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table. The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.
If n is the order of this reaction, then the rate equation is given by$$-\frac{d[AB]}{dt}=k[AB]^n$$If we take the log of both sides of this equation, we obtain:$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$. So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk. So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph? Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval. If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$The average concentration over this same time interval would be:$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table. The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.
Whenever you are given concentrations vs. time, think "integrated rate law." There are three integrated rate laws that your instructor will either give you or expect you to know:
(1) for a zero order reaction: $[A]_t = -kt + [A]_0$
(2) for a 1st order reaction: $ln[A]_t = -kt + ln[A]_0$
(3) for a 2nd order reaction: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$
An important thing to note is that each of these equations is in the same pattern as the equation for a line, viz., $y=mx+b$. For example, number 2 above (for the 1st order equation):
$\underbrace{ln[A]_t}_{y} = \underbrace{-kt}_{mx} + \underbrace{ln[A]_0}_{b}$
Why is this important? Because you can plot the data as expected by each of these equations and the one that plots as a straight line tells you the order of the reaction.
So, use your calculator or spreadsheet and enter the data. Then add two columns. One column computes the natural log of the given concentrations. The second column computes the reciprocal of the given concentrations.
Now create three separate plots. Plot "Time" on the horizontal axis of all three plots. Then, on the vertical axis of each separate chart, plot the given concentration in chart1, the natural log of concentration in chart2 and the reciprocal of concentration in chart3. You will find that chart3 is the graph of a straight line. Therefore, this is a second order reaction.
Your spreadsheet or calculator can calculate a linear regression line. The value of $k$ is the slope of that linear regression line. How do we know that? Because $k$ is the slope for each of the integrated rate law equations. In this case, the integrated rate law for this second order reaction is
$\underbrace{\frac{1}{[A]_t}}_{y} = \underbrace{kt}_{mx} + \underbrace{\frac{1}{[A]_0}}_{b}$
as shown above. (The $kt$ term is the $mx$ term in the equation for a line and $m$ is the slope.)
Finally, you can calculate the concentration of AB at 25 seconds using the equation for the regression line, which in this case is $y=0.0225x + 1.06$. Since we calculated the regression line for $\frac{1}{[AB]}$ we need to reverse that, so we want $[AB]_{25s} = \frac{1}{y}$:
$[AB]=\frac{1}{0.0225 \cdot 25 + 1.06}$ (the regression calculated the reciprocal so we need to reverse that)
$[AB]=0.616$
Whenever you are given concentrations vs. time, think "integrated rate law." There are three integrated rate laws that your instructor will either give you or expect you to know:
(1) for a zero order reaction: $[A]_t = -kt + [A]_0$
(2) for a 1st order reaction: $ln[A]_t = -kt + ln[A]_0$
(3) for a 2nd order reaction: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$
An important thing to note is that each of these equations is in the same pattern as the equation for a line, viz., $y=mx+b$. For example, number 2 above (for the 1st order equation):
$\underbrace{ln[A]_t}_{y} = \underbrace{-kt}_{mx} + \underbrace{ln[A]_0}_{b}$
Why is this important? Because you can plot the data as expected by each of these equations and the one that plots as a straight line tells you the order of the reaction.
So, use your calculator or spreadsheet and enter the data. Then add two columns. One column computes the natural log of the given concentrations. The second column computes the reciprocal of the given concentrations.
Now create three separate plots. Plot "Time" on the horizontal axis of all three plots. Then, on the vertical axis of each separate chart, plot the given concentration in chart1, the natural log of concentration in chart2 and the reciprocal of concentration in chart3. You will find that chart3 is the graph of a straight line. Therefore, this is a second order reaction.
Your spreadsheet or calculator can calculate a linear regression line. The value of $k$ is the slope of that linear regression line. How do we know that? Because $k$ is the slope for each of the integrated rate law equations. In this case, the integrated rate law for this second order reaction is
$\underbrace{\frac{1}{[A]_t}}_{y} = \underbrace{kt}_{mx} + \underbrace{\frac{1}{[A]_0}}_{b}$
as shown above. (The $kt$ term is the $mx$ term in the equation for a line and $m$ is the slope.)
Finally, you can calculate the concentration of AB at 25 seconds using the equation for the regression line, which in this case is $y=0.0225x + 1.06$. Since we calculated the regression line for $\frac{1}{[AB]}$ we need to reverse that, so we want $[AB]_{25s} = \frac{1}{y}$:
$[AB]=\frac{1}{0.0225 \cdot 25 + 1.06}$ (the regression calculated the reciprocal so we need to reverse that)
$[AB]=0.616$
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VOTE
If n is the order of this reaction, then the rate equation is given by $$-\frac{d[AB]}{dt}=k[AB]^n$$ If we take the log of both sides of this equation, we obtain: $$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$ This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$. So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk. So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph? Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval. If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as: $$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$ The average concentration over this same time interval would be: $$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$ So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table. The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.
If n is the order of this reaction, then the rate equation is given by$$-\frac{d[AB]}{dt}=k[AB]^n$$If we take the log of both sides of this equation, we obtain:$$\log\left(-\frac{d[AB]}{dt}\right)=n\log([AB])+\log k$$This equation is of the mathematical form $y=mx+b$ (i.e., a straight line), where $y=\log\left(-\frac{d[AB]}{dt}\right)$, $x=\log([AB])$, m = n, and $b=\log k$. So, if we could plot a graph of $y=\log\left(-\frac{d[AB]}{dt}\right)$ vs $x=\log([AB])$, we should get a straight line with a slope of m and and intercept of logk. So we could determine both the reaction order and the rate constant all in one shot from this single graph.
But how would we reduce the given data to plot such a graph? Well, we could use finite difference approximations for the average reaction rate and the average concentration over each time interval. If $[AB]_t$ represents the molar concentration of reactant AB at time t, we would approximate the average reaction rate over each time interval between two successive values of time $t_1$ and $t_2$ in the table as:$$\left(-\frac{d[AB]}{dt}\right)_{ave}=\frac{[AB]_{t_1}-[AB]_{t_2}}{t_2-t_1}$$The average concentration over this same time interval would be:$$[AB]_{ave}=\frac{[AB]_{t_1}+[AB]_{t_2}}{2}$$So we would plot the log of the average reaction rate as a function of the log of the average reactant concentration for all the time intervals in the table. The result should be a straight line with a slope equal to the reaction order and an intercept equal to the log of the reaction rate constant.
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VOTE