slope
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slope
How is it that the slope for both zero and first order reactions is -k but for second order reactions its k?
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Re: slope
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.
In conclusion, k is always positive and we only put a (-) sign in front to denote that the slope is negative.
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Re: slope
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.
Re: slope
[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)
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Re: slope
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!
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!
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Re: slope
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.
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Re: slope
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.
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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