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I just wanted to clarify the graphs associated with first order reactions. Since there is an ln in the graph, is there a curve to the graphs? Or is it still linear. Or are there two graphs one with exponential one without?
If you were to graph just concentration vs time, it would curve downwards, like an exponential decay function. However, if you graph ln of the concentration vs time, then it would be a straight line with a negative slope. This is because taking the log of an exponential is essentially cancelling out the exponential, making it just a value (which is the rate constant in our scenarios). If you can graph the ln of the concentration against time as a straight line, it is a first order reaction.
Dylan Mai 1D wrote:Will we ever need to graph on a test?
I went to Professor Lavelle's office hours and he quickly mentioned while explaining graphs that we won't be explicitly asked to graph. But, I think we should definitely have an understanding of what each of the graphs mean and be able to explain why they are curved/straight.
For a first order graph, the ln [A] vs. time graph will be linear and decreasing, since the concentration of the reactant is decreasing. If the graph was changed to represent [A] versus time, then the graph would be a decreasing exponential function.
The purpose of taking the ln of the concentration in the graph for a first order reactant is to have a straight-line graph instead of an exponential curve, as a linear graph is more useful for calculations. Similarly, the graph for second order reactants plots 1/[A] vs. time, and the graph of a zero order reaction plots [A] over time.
Josh Moy 1H wrote:A proper first order graph is just a straight decreasing line?
Yes, if you're graphing ln[A] vs time. All the "proper" graphs are going to look like straight lines, but the order depends on what's on the y-axis.
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