### Slope of 1st order RXNs

Posted:

**Sun Mar 11, 2018 7:37 pm**Is the slope of 1st order reactions +k or -k, and what is time graphed against in terms of [A]?

Created by Dr. Laurence Lavelle

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

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=148&t=29421

Page **1** of **1**

Posted: **Sun Mar 11, 2018 7:37 pm**

Is the slope of 1st order reactions +k or -k, and what is time graphed against in terms of [A]?

Posted: **Sun Mar 11, 2018 7:51 pm**

First order is -k slope with time vs ln[A]. Time vs just concentration [A] represents zero order (also -k slope).

Posted: **Sun Mar 11, 2018 8:49 pm**

Hey,

The slope of both 0 and 1st order reactions is -k, second order reactions have a positive k slope. Time is graphed against ln(a).

The slope of both 0 and 1st order reactions is -k, second order reactions have a positive k slope. Time is graphed against ln(a).

Posted: **Sun Mar 11, 2018 9:04 pm**

The integrated formula derived for the first order is [At] = [A0]*e^-kt. This gives an exponential graph of [At] against time and a decreasing graph with the slope of -k when graphed for ln[At] against time.

Posted: **Tue Mar 13, 2018 5:02 pm**

If the graph of time plotted against ln[A] is linear, then this indicates that the reaction with respect to reactant A is 1st order. The integrated rate law for 1st order reactions is ln[A] = -kt + ln[Ao]. If you look at this equation as if it is in the form of y = mx + b, you can see that the slope would be -k, not +k, due to the negative sign in front of kt. For zero order reactions, the slope is also -k for graphs of time plotted against [A]. However, the slope for 1/[A] vs. time for 2nd order reactions is +k.

Posted: **Tue Mar 13, 2018 5:03 pm**

The slope would be -k because the amount of reactants is decreasing over time.

Posted: **Thu Mar 15, 2018 8:03 pm**

First order integrated rate law: ln[A]= -k*t+ln[A]

y= m*X + B

Zero Order Integrated Rate Law: [A]=−kt+[A]initial

Y= m*X+B

y= m*X + B

Zero Order Integrated Rate Law: [A]=−kt+[A]initial

Y= m*X+B

Posted: **Sat Mar 17, 2018 10:49 am**

slope is -k

Posted: **Sat Mar 17, 2018 10:53 am**

remember only the second order has a positive k slope

Posted: **Sat Mar 17, 2018 11:43 am**

Slope is -K, a downward sloping graph.