# 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)#?

##### 1 Answer

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 *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?

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 *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")#

Based on these rates, *how much faster is the autoionization of water than the decomposition of* *?*