Here is the experimental data: Note all concentrations are the initial amounts

\begin{array}{cccc} \text{Experiment} & \ce{[C2H4Br2]} &\ce{[I-]} & \text{Initial rate of formation of }\ce{I3-}\\\hline 1 & 0.127 & 0.102 & 6.45\times 10^{-5}\\ 2 & 0.343 & 0.102 & 1.74\times 10^{-4}\\ 3 & 0.203 & 0.125 & 1.26\times 10^{-4}\\\hline \end{array}


We can determine the exponent $m$ easily because experiment one and two have the same initial $\ce{[I-]}$. However, when determining $n$, there is no data that has the same initial concentration. What the solution manual does is the following:

$$n= \dfrac {\ln \left(\dfrac{\mathrm{Rate}_{3}\cdot [\ce{C2H4Br2}]_2} {\mathrm{Rate}_{2} \cdot [\ce{C2H4Br2}]_3}\right)} {\ln \left(\dfrac{[\ce{I-}]_3}{[\ce{I-}]_2}\right)}=\dfrac{\ln \dfrac{(1.26\times 10^{-4})(0.343)}{(1.74\times 10^{-4})(0.203)}}{\ln \left(\dfrac{0.125}{0.102}\right)}=1$$

My question is, why are they multiplying the rate of the third experiment with the initial concentration from the second experiment and then multiply the rate from the second experiment with the initial concentration of the third experiment. I don't see any correlation with these numbers and how they can be used to determine the exponent. Any thoughts?


3 Answers 3


If you've correctly determined exponent $m$, then to ascertain $n$ you can proceed by first dividing rate equation $(3)$ by rate equation $(2)$ (note that the rate constant $k$ cancels immediately and we have that $m = 1$):

$$ \dfrac{\text{rate}_3}{\text{rate}_2} = \dfrac{\ce{[C2H4Br2]}_3}{\ce{[C2H4Br2]}_2} \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n $$

Rearranging this equation, one obtains:

$$ \dfrac{\text{rate}_3}{\text{rate}_2} \cdot \dfrac{\ce{[C2H4Br2]}_2}{\ce{[C2H4Br2]}_3} = \bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n $$

Now, to isolate $n$, take the logarithm (in any base of your choice, though the book arbitrarily uses the natural log, and I'll do likewise out of concern for consistency) of both sides of the equation, and remember that one of the basic properties of logarithms is that $\log_b x^n = n\log_b x$:

$$ \ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3} = \ln\bigg(\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}\bigg)^n = n\ln\dfrac{\ce{[I^{-}]}_3}{\ce{[I^{-}]}_2} $$

Solving for $n$, one finds:

$$ \frac{\ln\dfrac{\text{rate}_3 \ce{[C2H4Br2]}_2}{\text{rate}_2 \ce{[C2H4Br2]}_3}}{\ln\dfrac{\ce{[I-]}_3}{\ce{[I-]}_2}} = n $$

Which is the final equation from your textbook's solutions manual.


I think that there is a simple method to solve this question. I did this question with these 6 steps and my answer is right (although I had to do some approximation in calculation)

  1. Take values from experiment 1 and put them in rate equation. You will obatain an equation with $m$, $n$ and $k$ as varibales. Lets assume it equation 1.

  2. Now take values from experiment 2 and put them in rate equation. You obtain another eqaution like equation 1. Lets assume it equation 2.

  3. Divide equation 1 by 2 or 2 by 1 (whatever you like.) Solve it and you will easily get $m$ (equals to 1).

  4. Now put the value of $m$ in equation 2. You will obtain an equation with $n$ and $k$ as variables. Lets assume it equation 3.

  5. Now take values from experiment 3 and put them in rate equation along with value of $m$. You will get another equation like equation 3. Lets assume it equation 4.

  6. Divide equation 3 by 4 or 4 by 3 (whatever you like.) Solve it and you will easily get n (equals to 1).

PS, this might not be a 'by-the-book' method but I will solve it this way when I don't have too much time to solve a question and also because I am lazy enough to solve natural logarithm and stuff.


From the equation : $\text{rate}=k[\ce{C2H4Br2}]^m[\ce{I-}]^n$ you can get by applying logarithms : $\ln\text{rate}=\ln k + m\ln[\ce{C2H4Br2}]+n\ln[\ce{I-}]$

That has the form : $\mathrm Y= w_0+w_1x_1+w_2x_2 $ where $y=\ln\text{rate}$, $x_1=\ln[\ce{C2H4Br2}]$, $x_2=\ln[\ce{I-}]$ and $w_1=m$, $w_2=n$ and you can consider $x_0=1$ and $w_0=\ln k$ That becomes a linear regression problem, you can apply least squares or for better visualising it, you can draw the experimental values on logarithmic paper and fit a line. The result will be: $m=0.99884$ and $n = 0.98919$


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