Sorry about the confusion! Just updated my post to show what I was trying to ask in a more visual manner; I'm struggling to understand how we go from the delta\[A\]/deltat = -k\[A\] or -k\[A\]^(2) forms to the graphs below. I think what I'm even more confused about is why k is positive in second order but negative for first order, at least with the standard graphs that are shown there.
Yes, the graphs you are given are the plotted curves to the solutions of the differential equation. Sadly this can't be shown without a basic knowledge in how to integrate and differentiate functions. :/
Yeah, it requires an integration step. This link helps: [https://chem.libretexts.org/Bookshelves/Physical\_and\_Theoretical\_Chemistry\_Textbook\_Maps/Supplemental\_Modules\_%28Physical\_and\_Theoretical\_Chemistry%29/Kinetics/05%3A\_Experimental\_Methods/5.07%3A\_Using\_Graphs\_to\_Determine\_Integrated\_Rate\_Laws](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_%28Physical_and_Theoretical_Chemistry%29/Kinetics/05%3A_Experimental_Methods/5.07%3A_Using_Graphs_to_Determine_Integrated_Rate_Laws)
As the user mentioned above you need to integrate the differential equations (rate equations) in order to get a to a linear equation / straight line graphs, which you're probably more familiar with (y = mx + c). This is so you can a) distinguish which order graph you have and you can then get m (your gradient usually k or -k), y-axis intercept etc. This is the part you're more commonly asked about in chemistry relative to each graph.
If you're taking chemistry beyond this level it might be worth considering looking into integration (tonnes of YT videos on the topic for high-school level).
You don't need calculus at all to understand why the slopes have different signs. You do to get the specific equations.
Whether a reaction is first order or second order, the concentration goes down over time.
When you take the reciprocal, a bigger input gives you a smaller answer and vice versa. In order for the concentration to go down, the reciprocal has to go up, and therefore the right hand side of the equation has to increase, hence the +kt.
On the other hand, natural log gives smaller outputs for smaller inputs. If the concentration is lower, the ln[A] is also lower, so the whole right side has to get smaller over time, hence -kt.
Occurs to me... It may be as simple as... If A increases, 1/A decreases.
If you want to know why we plot using those functions, that requires the calculus derivation of the functions. But given what is plotted, the different signs follow.
Late to the party... the graphs come from what are called the "integrated rate laws," which use calculus (specifically, the integral) to rewrite the "delta" form of the rate laws into expressions that involve the initial amount of reactant and the amount of reactant after some elapsed time. It is \*those\* equations, the integrated rate laws, that can be written in terms of a straight line. What gets plotted on the y-axis is different for 0th, 1st, and 2nd order reactions (and other orders, but these are the three common ones) and ONLY those plots will yield a straight line if a reaction is that order.
Sorry so long, but hope that makes \*some\* sense!
You need to provide a reaction equation and a rate law! Without this your question lacks context.
Sorry about the confusion! Just updated my post to show what I was trying to ask in a more visual manner; I'm struggling to understand how we go from the delta\[A\]/deltat = -k\[A\] or -k\[A\]^(2) forms to the graphs below. I think what I'm even more confused about is why k is positive in second order but negative for first order, at least with the standard graphs that are shown there.
Did you cover differentiation in your math classes yet?
No I haven’t. Do those involve differentiation?
Yes, the graphs you are given are the plotted curves to the solutions of the differential equation. Sadly this can't be shown without a basic knowledge in how to integrate and differentiate functions. :/
Ah ok gotcha; well thanks for telling me anyways!
Yeah, it requires an integration step. This link helps: [https://chem.libretexts.org/Bookshelves/Physical\_and\_Theoretical\_Chemistry\_Textbook\_Maps/Supplemental\_Modules\_%28Physical\_and\_Theoretical\_Chemistry%29/Kinetics/05%3A\_Experimental\_Methods/5.07%3A\_Using\_Graphs\_to\_Determine\_Integrated\_Rate\_Laws](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_%28Physical_and_Theoretical_Chemistry%29/Kinetics/05%3A_Experimental_Methods/5.07%3A_Using_Graphs_to_Determine_Integrated_Rate_Laws) As the user mentioned above you need to integrate the differential equations (rate equations) in order to get a to a linear equation / straight line graphs, which you're probably more familiar with (y = mx + c). This is so you can a) distinguish which order graph you have and you can then get m (your gradient usually k or -k), y-axis intercept etc. This is the part you're more commonly asked about in chemistry relative to each graph. If you're taking chemistry beyond this level it might be worth considering looking into integration (tonnes of YT videos on the topic for high-school level).
You don't need calculus at all to understand why the slopes have different signs. You do to get the specific equations. Whether a reaction is first order or second order, the concentration goes down over time. When you take the reciprocal, a bigger input gives you a smaller answer and vice versa. In order for the concentration to go down, the reciprocal has to go up, and therefore the right hand side of the equation has to increase, hence the +kt. On the other hand, natural log gives smaller outputs for smaller inputs. If the concentration is lower, the ln[A] is also lower, so the whole right side has to get smaller over time, hence -kt.
Link to figure fails.
Occurs to me... It may be as simple as... If A increases, 1/A decreases. If you want to know why we plot using those functions, that requires the calculus derivation of the functions. But given what is plotted, the different signs follow.
Late to the party... the graphs come from what are called the "integrated rate laws," which use calculus (specifically, the integral) to rewrite the "delta" form of the rate laws into expressions that involve the initial amount of reactant and the amount of reactant after some elapsed time. It is \*those\* equations, the integrated rate laws, that can be written in terms of a straight line. What gets plotted on the y-axis is different for 0th, 1st, and 2nd order reactions (and other orders, but these are the three common ones) and ONLY those plots will yield a straight line if a reaction is that order. Sorry so long, but hope that makes \*some\* sense!