# Normal and heavy water mixing

This question seems silly, but this just came in my mind while i was learning about hydrogen.

Means will there be any reaction when $\ce{H2O}$ will be mixed with $\ce{D2O}$?

Yes, a reaction known as "hydrogen exchange" will take place. When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture. Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.
• Statistically, you would get that ratio, but unfortunately the $\Delta H_f$ for $D_2O$ is about 9 kJ/mol different from $H_2O$ so that the equilibrium is not that easy to determine. www1.lsbu.ac.uk/water/data.html – LDC3 Aug 16 '14 at 17:52