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### Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 7:27 pm
Why do we graph ln[A] against time instead of just [A] against time?

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 7:36 pm
You graph ln[A] vs time in order to get a linear graph, since it is ln[a] = -kt + c, not [a] = -kt + c. Using this you can find the value of k for the reaction, which would be the slope.

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 7:38 pm
We use ln[A] on the y-axis in order to get a linear graph. If you graph [A] vs t, you get a exponential decay. We use ln[A] to calculate the rate constant, -k, which is the slope on the line.

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 7:41 pm
ln[A] is used in order to yield a graph with a straight line, because otherwise it would yield a graph showing exponential decay. A graph with a straight line can be used to easily find the slope, which is equal to -k, giving us the equilibrium constant.

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 9:47 pm
For first order reactions, ln[A] is graphed against time for a linear graph which gives the slope. If just [A] were graphed against time, then the graph would be exponential.

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 10:13 pm
With a first order reaction graphing [A] vs t will produce a curve of exponential decay. By taking ln[A] we are able to make the graph a straight line.

### Re: Graphing 1st order reaction rates

Posted: Sun Mar 08, 2020 10:15 pm
because you want to look at the equation like an mx+b equation so that it's easier to graph. In this case, the ln[A] is the b.