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 $\ce{D2O}$?
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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 $\ce{D2O}$.
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1$\begingroup$ 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 $\endgroup$– LDC3Commented Aug 16, 2014 at 17:52
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$\begingroup$ @ LDC3 Excellent point LDC3, I'll edit my answer. $\endgroup$– ronCommented Aug 16, 2014 at 17:54
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$\begingroup$ @LDC3 the relevant quantity is not the difference vetween D2O and H2O, but the difference between their average and HDO. We would expect HDO to fall nearly halfway between H2O and D2O in enthalpy and free energybif formation, so the statistical model would likely be accurate in any fluid sample. $\endgroup$ Commented Jan 29 at 17:17