I'm really confused?

$$\ce{2H2O2 -> 2H2O + O2}$$

$$\begin{array}{ccc}\hline \text{Experiment no.} & c(\ce{H2O2}) & \text{rate} \\ \hline 1 & 14.45 & 3.35 \times 10^{-3}\\ 2 & 28.90 & 6.65 \times 10^{-3}\\ 3 & 57.80 & 13.35 \times 10^{-3}\\ \hline\end{array}$$

So from previous parts of the question, I found the order of reaction (1st), the rate equation for the reaction ($r = k[\ce{A}]$), the rate constant ($2.3 \times 10^{-4}\ \mathrm{s}^{-1}$, which is correct according to the answers), and now I need to find the rate of decomposition of $\ce{H2O2}$ when $\ce{[H2O2]}$ is exactly $10.0\ \mathrm{mmol\ L^{-1}}$.

I thought it's just $2.3 \times 10^{-4} \times 10.0$, which is $2.3 \times 10^{-3}$, but the answers provided is double that (so $4.6 \times 10 ^{-3}$), so I must have gotten the rate equation wrong?


It is a matter of definition:

The rate of the reaction is defined as: $$r=-\frac{1}{2}\frac{d[\ce{H2O2}]}{dt}$$ The rate of the decomposition of $\ce{H2O2}$ is defined as:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$

  • $\begingroup$ so if I'm calculating for k, would I use r = k[H2O2]? $\endgroup$ – Kris Jun 14 '16 at 14:38
  • $\begingroup$ Yes. Your calculation of k is quite right. $\endgroup$ – Yomen Atassi Jun 14 '16 at 14:39
  • $\begingroup$ so if, for example, the rate was r = (-2)*(d[H2O2]/dt), would rate of decomposition be (1/2)*r? am I understanding this relationship correctly? $\endgroup$ – Kris Jun 14 '16 at 14:48
  • $\begingroup$ In the expression of the reaction rate on devise by the stoichiometric coefficients of each reactant or product. $\endgroup$ – Yomen Atassi Jun 14 '16 at 16:03
  • $\begingroup$ The rate of decomposition of a reactant or the rate of formation of a product is given without dividing by the stoichiometric coefficients . $\endgroup$ – Yomen Atassi Jun 14 '16 at 16:05

The correct answer depends on what who wrote the problem had in mind about the values of rates in the last column in the data table:

If he considers the rate is rate of decomposition of $\ce{H2O2}$ , then the answer is $2.3\times 10^{−3}$

If he considers the rate is rate of formation of $\ce{O2}$, then the answer is $4.6\times10^{−3}$,

How we answer if we write the balanced equation in this way:

$\ce{H2O2 -> H2O + 1/2 O2}$


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