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Suppose we found that the reaction of Red#3 with bleach is 1st order in hypochlorite. The observed rate constant ($k_\mathrm{obs}$) for one of the trials was measured to be $\pu{0.00400 s^{-1}}$. If the concentration of hypochlorite was fixed at $\pu{0.134 M}$, calculate the specific rate constant ($k$) for that trial.

I am very confused in which equation I should use to solve this problem. Am I to use this equation? $$\mathrm{Rate} = k[A]^m[B]^n$$

If that is the case, for trial #1:

$$\mathrm{Rate} = \pu{0.00400 s^{-1}} \times \pu{0.134 M} = \pu{5.36e-4 mol L^{-1} s^{-1}}$$

But then what should I do next? This does not sound right. Please point out my mistakes.

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  • $\begingroup$ use the correct equation for first order kinetics. Yours does not fit. $\endgroup$
    – ssavec
    Feb 26, 2014 at 9:00

1 Answer 1

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We assume 1st order kinetics in hypochlorite and 1st order kinetics in the color. Putting this in the equation we get the following: $$\frac{\text{d}[\text{Red}]}{\text{d}t} = -k\, [\text{Red}]^1 \, [\text{bleach}]^1 $$

Because you keep the concentration of bleach the same throughout the trial, you can write $$k_\text{obs} = k\times [\text{bleach}]$$

Now, we can continue onwards as follows: $$ k_\text{obs} = k \times [\text{bleach}] \Leftrightarrow k = \frac{k_\text{obs}}{[\text{bleach}]} = 0.0299~\mathrm{s^{-1}\; M^{-1}}$$

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  • $\begingroup$ Thank you very much for the clear explanation! :) I get it now. $\endgroup$
    – Jesse
    Mar 2, 2014 at 2:23

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