Question #c4b96
1 Answer
The reaction is second order.
Explanation:
As you know, the rate of a reaction is simply a measure of how the concentrations or the partial pressures of the reactants (or of the products) change per unit of time.
The rate law of the reaction, which establishes a relationship between the rate constant,
#color(blue)("rate" = k * (P_a)^n)" "# , where
Now, the problems provides you with information about the rate of the reaction at two different partial pressure values.
You know that the rate of the reaction is equal to
#"rate"_1 = "1.07 torr s"^(-1) -># when#5%# of the reactant underwent decomposition
#"rate"_2 = "0.760 torr s"^(-1) -># when#20%# of the reactant underwent decomposition
Let's take
After
#P_1 = P_0 - 5/100 * P_0 = 95/100 * P_0#
Likewise, after
#P_2 = P_0 - 20/100 * P_0 = 80/100 * P_0#
You can now write
#"rate"_1 = k * (95/100 * P_0)^color(blue)(n)" "# and#" ""rate"_2 = k * (80/100 * P_0)^color(blue)(n)#
Divide these two equations to get rid of the rate constant
#"rate"_1/"rate"_2 = (color(red)(cancel(color(black)(k))) * (95/100 * P_0)^color(blue)(n))/(color(red)(cancel(color(black)(k))) * (80/100 * P_0)^color(blue)(n))#
This will be equivalent to
#"rate"_1/"rate"_2 = (95/100 * color(red)(cancel(color(black)(P_0))) * 100/80 * 1/color(red)(cancel(color(black)(P_0))))^color(blue)(n)#
#"rate"_1/"rate"_2 = (95/80)^color(blue)(n)#
Take the log from both sides to get
#log("rate"_1/"rate"_2) = log[ (95/80)^color(blue)(n)]#
#log("rate"_1/"rate"_2) = color(blue)(n) * log(95/80)#
Therefore,
#color(blue)(n) = log("rate"_1/"rate"_2)/log(95/80)#
Plug in your values to get
#color(blue)(n) = log((1.07 color(red)(cancel(color(black)("torr s"^(-1)))))/(0.760 color(red)(cancel(color(black)("torr s"^(-1))))))/(log(95/80)) = 1.991 ~~ color(green)(2) -># the reaction is second order