Calculate the rate of decomposition of H2O2 at 25 °C when [H2O2] is exactly 10.0 mmol/L I'm really confused? 

$$\ce{2H2O2 -&gt; 2H2O + O2}$$

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

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:$$r_{\mathrm{decomp}}=-\frac{d[\ce{H2O2}]}{dt}= 2r$$