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### 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]?

### Re: Slope of 1st order RXNs

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).

### Re: Slope of 1st order RXNs

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).

### Re: Slope of 1st order RXNs

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.

### Re: Slope of 1st order RXNs

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.

### Re: Slope of 1st order RXNs

Posted: Tue Mar 13, 2018 5:03 pm
The slope would be -k because the amount of reactants is decreasing over time.

### Re: Slope of 1st order RXNs

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

### Re: Slope of 1st order RXNs

Posted: Sat Mar 17, 2018 10:49 am
slope is -k

### Re: Slope of 1st order RXNs

Posted: Sat Mar 17, 2018 10:53 am
remember only the second order has a positive k slope

### Re: Slope of 1st order RXNs

Posted: Sat Mar 17, 2018 11:43 am
Slope is -K, a downward sloping graph.