Find the following for the reaction #"NO"(g) + "O"_3(g) -> "NO"_2(g) + "O"_2(g)#?
#a)# The reaction order overall, if the rate law is #k["NO"]["O"_3]# .
#b)# The units of the rate constant?
#c)# What happens to the rate if #["NO"]# doubles?
#d)# What happens to the rate if #["O"_3]# doubles?
1 Answer
The reaction order with respect to a reactant is just the exponent in the rate law. It is the degree of contribution the substance's concentration has on the initial rate.
Given the reaction order overall is the sum of the reaction orders with respect to each reactant, and the rate law...
#r(t) = k["NO"]["O"_3]# The overall order is
#1+1 = bb2# .
The units of this second-order reaction would be derived from the rate and concentration units in the rate law...
#r(t) = k["NO"]^1["O"_3]^1#
#"M"/"s" = (?)("M"^1)("M"^1)# Solve for
#(?)# , the units of#k# . Therefore, the units of#k# are:
#=> cancel"M"/"s" xx 1/("M"^cancel(2)) = 1/("M"cdot"s")# or#color(blue)("M"^(-1)cdot"s"^(-1))# .
The reaction is seen to be first order for EACH reactant. So if
#["NO"]# were doubled, the rate just doubles.
#color(red)(2)r(t) = k(color(red)(2)["NO"])^1["O"_3]^1# where
#r(t)# is the rate before doubling the concentration of#"NO"# .
Same thing for
#"O"_3# .
#color(red)(2)r(t) = k["NO"]^1 (color(red)(2)["O"_3])^1# where
#r(t)# is the rate before doubling the concentration of#"O"_3# .
