### zero order reaction

Posted:

**Sun Feb 14, 2016 12:14 pm**Why is it a zero order reaction with slope -k if the straight line fits to the data?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=144&t=12330

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Posted: **Sun Feb 14, 2016 12:14 pm**

Why is it a zero order reaction with slope -k if the straight line fits to the data?

Posted: **Sun Feb 14, 2016 1:56 pm**

Normally, with other order reactions, when we graph the change in concentration over time, the graph shows a curved line with a negative slope that shows the concentration first decreasing quickly and then slowing down. However, with zero order reactions, because the rate of the reaction is independent of [R], we end up with an equation for [A] that looks like this: [A]=-kt+[A]initial, which you'll notice looks exactly like the linear function mx+b, meaning that [A]initial is our y-intercept and -k is our slope.

Posted: **Sun Feb 14, 2016 6:42 pm**

Adding on to the above comment, zero-order reactions are independent of R because having a zero-order reaction means the rate is something like k*(concentration)^0 which equals k*1 which is simply a constant. The slope is going to be negative k because the concentration of the reactants is decreasing.