# Question #07e8f

##### 1 Answer

#### Answer:

The reaction is second-order.

#### Explanation:

You're dealing with a decomposition reaction of the form

#"AB " -> " products"#

for which the **differential rate law** can be written as

#"rate" = k * ["AB"]^n#

Here

#k# - therate constantfor the reaction

#n# - theorderof the reaction

According to experimental data, you have

Pick * any* two entries and plug the data into the differential rate law

#2.75 * 10^(-8)"mol dm"^(-3)"s"^(-1) = k * ("0.20 mol dm"^(-3))^n#

#11.0 * 10^(-8) "mol dm"^(-3)"s"^(-1) = k * ("0.40 mol dm"^(-3)")^n#

You can get rid of the unknown rate constant by dividing these two equations

#(2.73 * color(red)(cancel(color(black)(10^(-8)))) color(red)(cancel(color(black)("mol dm"^(-3)"s"^(-1)))))/(11.0 * color(red)(cancel(color(black)(10^(-8))))color(red)(cancel(color(black)("mol dm"^(-3)"s"^(-1))))) = color(red)(cancel(color(black)(k)))/color(red)(cancel(color(black)(k))) * (0.20)^n/(0.40)^n color(red)(cancel(color(black)(("mol dm"^(-3))^n)))/color(red)(cancel(color(black)(("mol dm"^(-3))^n)))#

#2.75/11.0 = (0.20/0.40)^n#

This is equivalent to

#ln(2.75/11.0) = ln[ (0.20/0.40)^n]#

Solve for

#ln(1/4) = n * ln(1/2)#

#ln[(1/2)^2] = n * ln(1/2)#

#2 * color(red)(cancel(color(black)(ln(1/2)))) = n * color(red)(cancel(color(black)(ln(1/2))))#

Therefore,

#n = 2#

and your decomposition reaction is **second-order**.

**SIDE NOTE** *The order of the reaction must come out the same regardless of what data you chose to use in the calculations.*

*I recommend using all the possible combinations of data sets to see that this is the case*.

#ln(24.75/11.0) = n * ln(0.60/0.40)#

#ln(2.75/24.75) = n * ln(0.20/0.60)#