Question

What is the rate of formation of water in its autoionization at `25^@ "C"` if its rate constant is `1.3 xx 10^(11) "L/mol"^(-1)"s"^(-1)`? What is the rate for dissociation of `"0.0010 M"` of `"N"_2"O"_5` if the rate constant is `1.0 xx 10^(-5) "s"^(-1)`?

Answer 1

Given the rate law

`r(t) = 1.3 xx 10^(11) "L/mol"^(-1)"s"^(-1)["OH"^(-)]["H"^(+)]`

and the usual concentrations of

`["OH"^(-)] = ["H"^(+)] = 10^(-7) "M"`

at `25^@ "C"` and `"1 atm"`, the rate is simply from plugging in and calculating.

`color(blue)(r(t)) = (1.3 xx 10^(11) "M"^(-1)"s"^(-1))(10^(-7) "M")^2`

`=` `color(blue)(1.3 xx 10^(-3) "M/s")`

This is a very fast reaction, as you might expect. The rate constant is huge. What do the units of the rate constant indicate for the order of the overall reaction?

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Given the rate law

`r(t) = k["N"_2"O"_5]`

for

`"N"_2"O"_5(g) rightleftharpoons "NO"_2(g) + "NO"_3(g)`,

and given the rate constant `k = 1.0 xx 10^(-5) "s"^(-1)` and initial concentration of `"N"_2"O"_5(g)`, the initial rate is again, just plugging in and using the equation.

`color(blue)(r(t)) = (1.0 xx 10^(-5) "s"^(-1))("0.0010 M") = color(blue)(1.0 xx 10^(-8) "M/s")`

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Based on these rates, how much faster is the autoionization of water than the decomposition of `"N"_2"O"_5`?