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Why the reaction rate (nitration) of $\ce{C6H6}$, $\ce{C6D6}$ and $\ce{C6T6}$ are equal? Can anyone explain this using simple words?

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    $\begingroup$ I suspect the rates seem to be the same within an error of measurement. $\endgroup$ – andselisk Nov 8 '17 at 5:33
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    $\begingroup$ If you measured the rates carefully, you would probably see an inverse kinetic isotope effect (that is, the reaction will very slightly speed up as one goes from $\ce{C6H6}$ to $\ce{C6T6}$) because nitration involves an $sp^2$ to $sp^3$ change in the rate determining step. $\endgroup$ – levineds Nov 8 '17 at 7:03
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    $\begingroup$ In simple words, H, D, and T are isotopes*, and from chemical point of view they are just the same. Only if you measure them with really, really great precision, you may notice some differences. $\endgroup$ – Ivan Neretin Nov 8 '17 at 8:04
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The rate-determining step is the attack of the benzene aromatic ring onto the electrophile, and the loss of aromaticity is what causes the instability of the transition state, resulting in a higher activation energy and thus a slower rate of reaction. The subsequent deprotonation is after the rate-determining step and thus will not affect the rate of reaction, despite C-T > C-D > C-H bond strength.

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In nitration of benzene or $\ce{C6D6}$, the rate determining or slowest step is the formation of the nitronium ion $\ce{NO2+}$ and not the formation of sigma complex as in the other electrophilic aromatic substitutions, therefore the rate is unaffected by the leaving of a hydrogen or deuterium atom.

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