slope


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Miranda 1J
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Joined: Fri Sep 29, 2017 7:06 am

slope

Postby Miranda 1J » Thu Mar 01, 2018 1:08 am

How is it that the slope for both zero and first order reactions is -k but for second order reactions its k?

Hannah Chew 2A
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Re: slope

Postby Hannah Chew 2A » Thu Mar 01, 2018 9:25 am

Since k, the constant, is always positive, we must change the sign in front of k to denote the slope. So for zero order reactions, if we plot [A] vs time, the slope is (-), meaning that slope = (-)k. For first order reactions, if plot ln[A] vs time, the slope is also (-), so we set that equal to (-)k. Lastly, for second order reactions, 1/[A] vs time yields a positive slope, so we can just leave the slope = k.

In conclusion, k is always positive and we only put a (-) sign in front to denote that the slope is negative.

Erica Nagase 1H
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Joined: Sat Jul 22, 2017 3:00 am

Re: slope

Postby Erica Nagase 1H » Thu Mar 01, 2018 10:19 am

For a zero order reaction, the graph of [A] vs time is negative because the reactant concentration is decreasing as the reaction proceeds. For a first order reaction, the graph of ln[A] vs time is negative because as [A] decreases, ln[A] also decreases. For a second order reaction, a graph of 1/[A] vs time is positive because as [A] decreases, the value of 1/[A] will increase.

Sophie 1I
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Joined: Fri Sep 29, 2017 7:04 am

Re: slope

Postby Sophie 1I » Fri Mar 02, 2018 3:20 pm

Reactant concentration is decreasing so the slope will be negative so therefore k must be negative.
Second order reactions of 1/[A] has a positive slope so k is positive

404995677
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Joined: Fri Sep 29, 2017 7:07 am

Re: slope

Postby 404995677 » Fri Mar 02, 2018 5:18 pm

[A] is the concentration of the reactants as the reaction continues. For the forward reaction, the concentration of the reactants decrease, and for second order reactions, the graph is graphed 1/[A] vs. time. So if [A] is decreasing over time, the graph will show a line/slope that is increasing (or a k that is positive) because 1/[A] will increase as [A] decreases. (ex: 1/[1]=1, and 1/[0.001]=1,000)

Kevin Tabibian 1A
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Re: slope

Postby Kevin Tabibian 1A » Sat Mar 03, 2018 5:31 pm

Whatever it takes to make k positive

204932558
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Re: slope

Postby 204932558 » Sat Mar 03, 2018 8:51 pm

The graph for zero order reaction and first order reaction is {A} against time. But for second order reaction it's 1/[A] against time, which is why the slope is negative.

Sabrina Fardeheb 2B
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Re: slope

Postby Sabrina Fardeheb 2B » Sun Mar 04, 2018 5:06 pm

You can plot the concentration values and examine the straight (it should be straight if it's an order reaction I believe) line to see what the value of the slope is.

In addition, you can derive the integrated rate laws and use the equation y = mx + b to look at the sign.

For example, for the zero order reaction, the integrated rate law is [A]t = -kt + [A]0
It can also be written as y = - (slope)(x) + intercept. Because there is a negative sign in front of the k, you should expect a negative slope. This also works for the integrated rate law for the second order reaction! Hope this helps!

Belle Calforda3f
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Re: slope

Postby Belle Calforda3f » Sun Mar 04, 2018 5:12 pm

Taking the integral when solving for the integrated rate law for a second order reaction produces a double negative which turns into a positive. Integrating the integrated rate law for each order will demonstrate this.

Ashley Garcia 2L
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

Re: slope

Postby Ashley Garcia 2L » Mon Mar 05, 2018 4:17 pm

Using A → B

Zero order reaction: graph [A] vs. time, this plot will produce a negative slope. Therefore, slope=-k because the rate constant k must be a positive value.
1st order reaction: graph ln[A] vs. time, this plot will produce a negative slope. Therefore, slope=-k because the rate constant k must be a positive value.
2nd order reaction: graph 1/[A] vs. time, this plot will produce a positive slope. Therefore, slope=k because the rate constant k must be a positive value.


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