# Reaction rate constant conversion [closed]

I'm flustered, I'm not into chemistry and need to convert $\mathrm{(nM\cdot s)^{-1}}$ to $\mathrm{(M\cdot s)^{-1}}$. Is it correct to multiply with $10^9$?

## closed as off-topic by paracetamol, Todd Minehardt, M.A.R. ಠ_ಠ, hBy2Py, airhuffApr 25 '17 at 15:36

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• Hi and welcome to chemistry.stackexchange.com. Feel free to take a tour of the site. I am unsure what your question is supposed to be; turning nano-molar-seconds into mols? That doesn’t really make sense to me … – Jan Nov 18 '15 at 9:29
• I'm designing a model of the EGFR pathway with simmune-software-tool and have to extract all kind of information from journals/papers about association rates and dissociation rates of the proteins in the pathway. – unterm- Nov 18 '15 at 10:02
• Wouldn't the conversion of asssumed nM into M imply dividing by $10^9$ instead of the multiplication mentioned in the OP? Alternatively, a multiplication with $10^{-9}$ instead of $10^{9}$ ... – Buttonwood Apr 25 '17 at 12:08

You have a rate constant for a second order reaction. You want to convert from $(\mathrm{nM}^{-1} \cdot s^{-1})$ to $(\mathrm{M}^{-1} \cdot s^{-1})$. So you have to write it out using fractions, like
$$\mathrm{nM}^{-1} s^{-1} = \frac{1}{\mathrm{nM} \cdot \mathrm{s}}\times \frac{10^9 \mathrm{nM}}{\mathrm{M}} = 10^9 \mathrm{M}^{-1} s^{-1}$$
So yes, you multiply the rate by $10^9$