# 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)?

Apr 18, 2017

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 ${25}^{\circ} \text{C}$ and $\text{1 atm}$, the rate is simply from plugging in and calculating.

$\textcolor{b l u e}{r \left(t\right)} = {\left(1.3 \times {10}^{11} \text{M"^(-1)"s"^(-1))(10^(-7) "M}\right)}^{2}$

$=$ $\textcolor{b l u e}{1.3 \times {10}^{- 3} \text{M/s}}$

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?

Given the rate law

$r \left(t\right) = k \left[{\text{N"_2"O}}_{5}\right]$

for

${\text{N"_2"O"_5(g) rightleftharpoons "NO"_2(g) + "NO}}_{3} \left(g\right)$,

and given the rate constant $k = 1.0 \times {10}^{- 5} {\text{s}}^{- 1}$ and initial concentration of ${\text{N"_2"O}}_{5} \left(g\right)$, the initial rate is again, just plugging in and using the equation.

$\textcolor{b l u e}{r \left(t\right)} = \left(1.0 \times {10}^{- 5} \text{s"^(-1))("0.0010 M") = color(blue)(1.0 xx 10^(-8) "M/s}\right)$

Based on these rates, how much faster is the autoionization of water than the decomposition of ${\text{N"_2"O}}_{5}$?