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
#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
#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
