graph for 1st, 2nd, and zero order reactions
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graph for 1st, 2nd, and zero order reactions
What would the graphs for the 1st, 2nd, and zero order reactions look like?
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Re: graph for 1st, 2nd, and zero order reactions
In order to find out whether or not a reactant is zero, first, or second-order based on experimentally derived data, you would have to plot different values against time to see which one most strongly resembles a straight or linear line. For a reactant to be considered zero-order, plotting its concentration against time ([A] vs t) should result in a decreasing linear line. For a reactant to be considered first order, plotting the natural log against time (ln([A]) vs t) should result in a decreasing linear line. Finally, for a reactant to be considered second-order, plotting the inverse of its concentration (1/[A] vs t) should result in an increasing linear line.
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Re: graph for 1st, 2nd, and zero order reactions
A zero order reaction will have a linear plot (straight line) when you plot [A] versus time.
First order will have a linear graph when you plot ln[A] versus time.
Second order will have a linear graph when you plot 1/[A] versus time (this will have a positive slope unlike the rest).
[A] is the concentration of the reactant.
First order will have a linear graph when you plot ln[A] versus time.
Second order will have a linear graph when you plot 1/[A] versus time (this will have a positive slope unlike the rest).
[A] is the concentration of the reactant.
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Re: graph for 1st, 2nd, and zero order reactions
From Friday's lecture, here's a visual of the graphs Dr. Lavelle drew on the whiteboard : hope this helps :)
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Re: graph for 1st, 2nd, and zero order reactions
The graph of a zero-order reaction would be linear with a negative slope, this is because the rate law is: rate = -k[A]
The graph of a first-order reaction would be linear with a negative slope, this is because the rate law is: rate = -kln[A]
- for a natural log, the smaller the concentration the faster the rate so it makes sense that the slope is negative
The graph of a second-order reaction would be linear with a positive slope, this is because the rate law is: rate = k/[A]
- since the concentration of the reactant is on the denominator
The graph of a first-order reaction would be linear with a negative slope, this is because the rate law is: rate = -kln[A]
- for a natural log, the smaller the concentration the faster the rate so it makes sense that the slope is negative
The graph of a second-order reaction would be linear with a positive slope, this is because the rate law is: rate = k/[A]
- since the concentration of the reactant is on the denominator
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Re: graph for 1st, 2nd, and zero order reactions
Different functions graphed will give a linear curve. From this, you can determine what order the reaction is.
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Re: graph for 1st, 2nd, and zero order reactions
To figure out the order, you would have to plot the data against time and figure out which is the most linear. A zero order reaction would be linear with a negative slope. A first order reaction would be linear with a negative slope. A second order reaction would be linear with a positive slope.
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Re: graph for 1st, 2nd, and zero order reactions
Like some of the other replies here, the first and zero order reactions have negative slopes, but the second order reaction has a positive slope! Hope this helps
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