## Slope of 1st order RXNs

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Sophia Bozone 2G
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### Slope of 1st order RXNs

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

Zane Mills 1E
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### Re: Slope of 1st order RXNs

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

Shawn Patel 1I
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### Re: Slope of 1st order RXNs

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).

Niyanta Joshi 1F
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### Re: Slope of 1st order RXNs

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.

Michelle Lu 1F
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### Re: Slope of 1st order RXNs

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.

Wayland Leung
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### Re: Slope of 1st order RXNs

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

Angel R Morales Dis1G
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### Re: Slope of 1st order RXNs

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

Sarah 2F
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### Re: Slope of 1st order RXNs

slope is -k

Nicole Jacobs 1C
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### Re: Slope of 1st order RXNs

remember only the second order has a positive k slope

Cynthia Bui 2H
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### Re: Slope of 1st order RXNs

Slope is -K, a downward sloping graph.