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Integrated Rate Laws

An integrated rate law is a rate law that relates concentration to time. These are used often when we need to know the specific concentration of a reactant at a certain time. The integrated rate laws are derived directly from the differential rate laws using calculus, which will be shown in the fun facts section.



First Order Integrated Rate Law

The first order integrated rate law is:


`"ln"[A] = -kt + "ln"[A]_0`


Where `[A]_0` is the initial concentration.


Notice that if we substitute `"ln"[A]=y` and `"ln"[A]_0=b` , we can write the first order integrated rate law as `y=-kt+b` which is the formula for a line from algebra (remember `y=mx+b` ). In this case, the slope is `-k` . This means that if we were to make a graph with `"ln"[A]` on the y-axis and `t` on the x-axis, we can graph the equation



Second Order Integrated Rate Law

The second order integrated rate law is:


`1/[A]=kt + 1/([A]_0)`


Just like with the first order integrated, we can make a linear graph if we plot `1/[A]` on the y-axis and `t` on the x-axis



Zeroeth Order Integrated Rate Law

The zeroeth order integrated rate law is:


`[A]=-kt+[A]_0`


A linear graph can be made by plotting `[A]` on the y-axis and `t` on the x-axis.



Summary

1. An integrated rate law relates concentration to time.

2. Each of the rate laws can be plotted as a linear graph with `k` as the slope.



Derivations

NOTE: requires knowledge of calculus and differential equations.

#1. First Order Integrated Rate Law


`"rate"=(Delta[A])/(Deltat)=(d[A])/(dt)=-k[A]`


`(d[A])/(dt)=-k[A]`


`(d[A])/[A]=-k dt`


`∫ (d[A])/[A]=∫ kdt`


`"ln"[A]=-kt + C`


At `t=0` , `[A]` is the initial concentration `[A]_0` .[


`"ln"[A]_0=0 + C`


`C="ln"[A]_0`


`∴"ln"[A]=-kt + "ln"[A]_0`


#2. Second Order Integrated Rate Law


`"rate"=(Delta[A])/(Deltat)=(d[A])/(dt)=-k[A]^2`


`(d[A])/(dt)=-k[A]^2`


`(d[A])/([A]^2)=-kdt`


`∫ (d[A])/([A]^2)=∫ -kdt`


`1/[A]=kt+C`


At `t=0` , `[A]=[A]_0`


`1/([A]_0)=0+C`


`C=1/([A]_0)`


`∴1/[A]=kt+1/([A]_0)`



#3. Zeroeth Order Integrated Rate Law


`"rate"=(Delta[A])/(Deltat)=(d[A])/(dt)=-k[A]^0`


`(d[A])/(dt)=-k[A]^0`


Since anything raised to the power of `0` is `1` , `[A]^0=1`


`(d[A])/(dt)=-k`


`∫ d[A]=∫ -kdt`


`[A]=-kt + C`


At `t=0` , `[A]=[A]_0`


`[A]_0=0+C`


`C=[A]_0`


`∴[A]=-kt + [A]_0`