5
$\begingroup$

Why does the graph of $\ln[\ce A]$ vs $t$ produce a straight line for the first order rate law?

Also why does the graph of $t$ vs $\frac{1}{B}$ produce a straight line for the second order rate law?

$\endgroup$
9
$\begingroup$

The short answer is that the equations you describe as graphs are the solutions of the differential equations that define the rate laws in question. If you don't understand what a differential equation is, any sensible answer is going to be problematic.

Assuming you do know what a differential equation is, we proceed!

For a first order rate law, the differential equation is something like:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$

You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:

$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

Similarly, the differential equation for the second order rate law is:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$

And again integrate to get:

$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

This all comes from the following Wikipedia page, which has a lot more details: Rate equation

$\endgroup$
0
0
$\begingroup$

]Wikipedia got it wrong for the 2nd order differential rate equation. It should be this:

-d[A] / dt = k[A]^2 for a second order reaction, where there is no 2 by the k constant, and the negative sign denotes the disappearance of the concentration of A with respect to time, and where d[A] / dt means the derivative of the concentration of [A] with respect to time (dt).

Divide both [A]^2 and multiply by negative dt:

{1 / [A]^2} d[A] = -k dt

[A]^(-2) d[A] = -k dt ... This is a separable equation.(Actually, 0 order to third order and even three-halves order are separable differential equations.)

Integrate both sides with definite bounds.

(from [A]0 to [A]t ∫[A]^(-2) d[A] = -k (from 0 to t) ∫(dt),

use power rule of integration (only the first law is special because if you use power rule of integration on 1 / [A] you get a divide by zero error.) So in power rule of integration, you add 1 to the existing power which is -2 and then divide by the new power which is -2 + 1 = -1, so that gives us the integration [A]^(-1) / -1 or -1 / [A]. And when you integrate dt, you are really integrating t^0 dt, so you get t, and "k" is a constant (non variable) so it is not integrated ever in all the rate laws.

-(1 / [A]) {[from [A]0 to [A]t} = -kt {from t = 0 to t} Since both sides are negative, divide both sides of the equation by -1, and everything's positive.

(1 / [A]) {[from [A]0 to [A]t} = kt {from t = 0 to t}

(1 / [A]t) - (1 / [A]0) = k(t - 0) = k(t) Then some algebra, adding 1 / [A]0 to both sides and tada!!

(1 / [A]t) = (1 / [A]0) + k(t)

Now how do we graph this linear? think about it. If we substitute 1 / [A]t as "y" and (1 / [A]0) as "b" and k(t) is m(x), that's basically the equation of a line: y = m(x) + b.

So you redefine the axes with the vertical axis as 1 / [A], and the horizontal axis as time (t), and you expect a straight line if your chemical reaction is second order.

Theoretically, you should be able tabulate your data and calculate logarithms and inverses and make tables, and compare how well the scatter plots form a line to figure out which model is the best model. But, if you do not follow the reaction over enough time, then all the models could have good fits which is undesirable.

Not showing the work, but for the 1st order reaction, the integrated rate law is this:

ln[A]t = ln[A]0 - k(t),

where remember [A]0 is the conc of reactant A at time 0, and [A]t is the concentration of reactant A at any given time. So if I just replace the ln[A] as the units of the vertical axis, and the "t" as the units of the horizontal axis, then I get another y = m(x) + b line.

Once you get your axes right in a table, you just format, you just do a linear regression in Microsoft Excel, QtiPlot, R, etc... to get the equation of your line to fit the model.

I hope I helped you understand this.

Here are some rate laws in all their forms

Here is how you integrate the first order rate law: https://www.youtube.com/watch?v=smcyZ_r0A88

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.