# 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

**order of the reaction**

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 reactantunderwent decomposition

#"rate"_2 = "0.760 torr s"^(-1) -># when#20%# of the reactantunderwent decomposition

Let's take

After **undergoes decomposition**, you will be left with

#P_1 = P_0 - 5/100 * P_0 = 95/100 * P_0#

Likewise, after **undergoes decomposition**, you will be left with

#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 issecond order