Rate laws relate rate and concentration of reactants. They also relate the concentration of reactants with time. These are called** integrated rate laws.**

Integrated rate laws are derived from differential rate laws. For any chemical reaction, differential rate laws can be integrated with respect to time to give an equation that relates the concentration of reactants or products with the time of reaction.

Now, we study integrated rate laws for zero, first, 2nd, and 3rd order reactions one by one.

# For zero-order reaction

The reaction in which the rate is independent of the concentration of reactants is called a zero-order reaction.

We make a graph to describe the change in concentration of reactants or products with time from the set of data of concentration vs time.

Integrated rate equations for the zero-order reaction is

_{t}=-k_{o}t+_{o}→ Eq(1)

Where,

_{t}=concentration of reactants at time ‘t’ sec

_{o}= concentration of reactants at time=0 sec

k_{o}=rate constant for zero-order reaction

_{t}=-k_{o}t+_{o}→ Eq(1)

**Question. How many ways can the equation be plotted?**

_{t }vs time

From eq(1)

_{t}=-k_{o}t+_{o}

Compare eq(1) with the straight-line equation **y=mx+c** and plot a graph between _{t }on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

This graph shows that the concentration of reactants is decreasing with time. So the graph is a straight line with a negative slope -k_{o} and intercepts _{o}.

**d vs time**

From eq(1)

_{t}=-k_{o}t+_{o}

_{t}–_{o}=-k_{o}t

**d=-k _{o}t→Eq(2)**

Compare eq(2) with the equation of straight line** y=mx+c** and plot a graph between don the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph shows a decreasing slope because changes in concentration of reactants are decreasing with time. It has slope -k_{o} and the intercept is zero.

**-d vs time**

From eq(2)

d=-k_{o}t

Multiply both sides of the equation with a negative sign, we get

**-d=k _{o}t→Eq(3)**

Compare eq(3) with the equation of straight line **y=mx+c** and plot a graph between **-d**on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph shows a direct relation between decreasing change in concentration of reactants with time. The graph is a straight line with a positive slope and the intercept is zero.

**vs k**_{o}

-d=k_{o}t→Eq(3)

Divide L.H.S of eq(3) with time.

**=-k _{o}→Eq(4)**

Compare eq(4) with the equation of straight line** y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

Eq(4) is not an equation of a straight line. So the plot is not formed.

### Unit of rate constant

The unit of rate constant for a zero-order reaction is **mol sec ^{-1}.**

### Half-life

Half-life is a time in which half of the reactants are converted into products.

The half-life for a zero-order reaction is

**t _{1/2}=**

## For 1st order reaction

The reaction in which the rate depends on the first power of concentration of reactants is called a first-order reaction.

To describe the change in concentration of reactants or products with time we make different plots.

Integrated rate equations for the 1st-order reaction is

**ln _{t}=-k_{1}t+ln_{o}Eq(1)**

Where,

_{t}=conc of reactants at time=t sec

_{o}=conc of reactants at t=0sec

k_{1}=rate constant for 1st order reaction

**Question. How many ways can the equation be plotted?**

**ln**_{t}vs time

Compare eq(1) with the equation of straight line** y=mx+c** and plot a graph between ln** _{t }**on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph shows that the concentration of reactants decreases with time so it a straight-line graph with a negative slope m=-k_{1} and intercept c=ln_{o}.

**ln**_{o}vs time

From eq(1)

ln_{t}=-k_{1}t+ln_{o}

**ln _{o}=k_{1}t+ln_{t}Eq(2)**

Compare eq(2) with the equation of straight line** y=mx+c** and plot a graph between ln** _{o }**on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

_{o }is the concentration of reactants when time t=0 sec. It means it is not changing with time. So the graph is parallel to the x-axis and slope m=0.

**ln vs time**

From eq(1)

ln_{t}=-k_{1}t+ln_{o}

ln_{t}-ln_{o}=-k_{1}t

**=-k _{1}tEq(3)**

Compare eq(3) with the equation of straight line **y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{1} and zero intercept c=0.

_{t}vs time

From eq(3)

=-k_{1}t

=e^{-k1t}

_{t}=_{o}e^{-k1t}Eq(4)

Compare eq(4) with the equation of straight line **y=mx+c **and plot a graph between _{t} on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

This graph shows an exponential decrease.

_{o}–_{t}vs time

Now plot a graph between ** _{o}**–

**on the y-axis and time on the x-axis.**

_{t }**Graph **

## Explanation of graph

** _{o}**–

**is the concentration of the product.it has a linear relationship with time. So the graph is a straight line.**

_{t }### Unit of rate constant

Unit of rate constant for 1st order reaction is **sec ^{-1}.**

### Half-life

Half-life is a time in which half of the reactants are converted into products.

The half-life for 1st order reaction is

**t _{1/2}=**

# For 2nd order reaction

The reaction in which the rate depends on the second power of concentration of reactants is called a second-order reaction.

To describe the change in concentration of reactants or products with time we make different plots.

In the 2nd order reaction, there are two cases.

- 2nd order reaction with the same initial concentrations.
- 2nd order reaction with the different initial concentration

The integrated rate equation for second-order reaction with the same concentration of reactants

**=+k _{2}tEq(1)**

**Question. How many ways can the equation be plotted?**

**vs time**

From eq(1)

**=+k _{2}t**

Compare equation(1) with the equation of straight line **y=x+mc** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=k_{2} and intercept c=.

**vs time**

From eq(1)

=+k_{2}t

-=k_{2}t

**=k _{2}tEq(2)**

Compare eq(2) with the equation of straight line** y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with positive slope m=k_{2} and zero intercept c=0.

The integrated rate equation for 2nd order reaction with different concentrations of reactants

**ln=k _{2}t+lnEq(3)**

**Question. How many ways can the equation be plotted?**

**lnvs time**

From eq(3)

ln=k_{2}t+ln

Compare eq(3) with the equation of straight line** y=mx+c **and plot a graph between lnon the y-axis and time on the x-axis.

** Graph **

## Explanation of graph

The graph is a straight line with positive slope m=k_{2 }and intercept c=ln.

**lnvs time**

From eq(3)

ln=k_{2}t+ln

Taking negative signs common from (b-a)

**ln=k _{2}t+lnEq(4)**

Compare eq(4) with the equation of straight line **y=mx+c** and plot a graph between lnon the y-axis and time on the x-axis.

** Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=k_{2 }and intercept c=ln.

**lnvs time**

From eq(3)

ln=k_{2}t+ln

ln=k_{2}t+ln

Multiply both sides with a negative sign

ln=-k_{2}t-ln

**ln=-k _{2}t+lnEq(5)**

Compare eq(5) with the equation of straight line** y=mx+c **and plot a graph between lnon the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{2 }and intercept c=ln.

**lnvs t**

From eq(3)

ln=k_{2}t+ln

Invert the term we get,

ln()=k_{2}t+ln()

ln()=-k_{2}t-ln()

Change the intercept corresponding to y

**ln()=-k _{2}t+ln()Eq(6)**

Compare eq(6) with the equation of straight line **y=mx+c** and plot a graph between ln() on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{2 }and intercept c=ln.

**ln() vs t**

From eq(3)

ln=k_{2}t+ln

Invert the term we get,

ln()=k_{2}t+ln()

ln()=k_{2}t+ln()

Change the intercept corresponding to y

**ln()=k _{2}t+ln()Eq(7)**

Compare eq(7) with the equation of straight line y=mx+c and plot a graph between ln() on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=k_{2 }and intercept c=ln.

**ln() vs t**

From eq(3)

ln=k_{2}t+ln

ln()-ln()=k_{2}t

Taking common

=k_{2}t

ln=k_{2}t

Apply property of log

**ln=k _{2}tEq(8)**

Compare eq(8) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with positive slope m=k_{2 }and intercept c=0.

**ln() vs t**

From eq(8)

ln=k_{2}t

ln=k_{2}t

**ln=-k _{2}tEq(9)**

Compare eq(9) with the equation of straight line** y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{2 }and intercept c=0.

**ln() vs t**

From eq(8)

ln=k_{2}t

Invert the term

ln=k_{2}t

**ln=-k _{2}tEq(10)**

Compare eq(10) with the equation of straight line **y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{2 }and intercept c=0.

**ln() vs t**

From eq(8)

ln=k_{2}t

Invert the term

ln=k_{2}t

**ln=k _{2}tEq(11)**

Compare eq(11) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=k_{2 }and intercept c=0.

**ln() vs t**

From eq(8)

ln=k_{2}t

ln=k_{2}t

Invert the term

ln=k_{2}t

Invert the term

ln=k_{2}t

**ln=-k _{2}tEq(12)**

Compare eq(12) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

The graph is a straight line with a negative slope m=-k_{2 }and intercept c=0.

**ln() vs t**

From eq(8)

ln=k_{2}t

ln=k_{2}t

Invert the term

**ln=k _{2}tEq(13)**

Compare eq(13) with the equation of straight line **y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

The graph is a straight line with a positive slope m=k_{2 }and intercept c=0.

**ln vs t**

From eq(3)

ln()=k_{2}t+ln

ln(-k_{2}t=ln

ln(-k_{2}t=ln

ln()+k_{2}t=ln

ln()+k_{2}t=ln

**ln=ln+k _{2}tEq(14)**

Compare eq(14) with the equation of straight line **y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

Since is a constant term, so it is not changing with time. The graph is a straight line parallel to the x-axis and slope m=0.

### Unit of rate constant

Unit of rate constant for the 2nd order reaction is** M ^{-1} sec^{-1}.**

### Half-life

### Half-life is a time in which half of the reactants are converted into products.

The half-life for 2nd order reaction is

**t _{1/2}=**

# For 3rd order reaction

The reaction in which the rate depends on the third power of concentration of reactants is called a third-order reaction.

To describe the change in concentration of reactants or products with time we make different plots.

Here we discuss only one case when initial concentrations of all reactants are the same.

The integrated rate equation for 3rd order reaction is

**=2k _{3}t+Eq(1)**

**Question. How many ways can the equation be plotted?**

**Vs time**

From eq(1)

=2k_{3}t+

Compare eq(1) with the equation of straight line y=mx+c and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=2k_{3 }and intercept c=.

**Vs time**

From eq(1)

=2k_{3}t+

-=2k_{3}t

=2k_{3}t

By solving, we get

**=2k _{3}tEq(2)**

Compare eq(2) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with a positive slope m=2k_{3 }and intercept c=0.

**Vs time**

From eq(2)

=2k_{3}t

Taking negative sign common

=2k_{3}t

Multiply both sides with a negative sign

**=-2k _{3}tEq(3)**

Compare eq(3) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with negative slope m=-2k_{3 }and intercept c=0.

**Vs time**

From eq(1)

=2k_{3}t+

-=2k_{3}t

=2k_{3}t

Taking negative sign common

=2k_{3}t

Multiply both sides with a negative sign

**=-2k _{3}tEq(4)**

Compare eq(4) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph **

## Explanation of graph

The graph is a straight line with negative slope m=-2k_{3 }and intercept c=0.

**Vs time**

From eq(1)

=2k_{3}t+

-=2k_{3}t

**=2k _{3}tEq(5)**

Compare eq(5) with the equation of straight line **y=mx+c **and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

The graph is a straight line with a positive slope m=2k_{3 }and intercept c=0.

**Vs time**

From eq(1)

=2k_{3}t+

-2k_{3}t=

**=-2k _{3}tEq(6)**

Compare eq(6) with the equation of straight line **y=mx+c** and plot a graph between on the y-axis and time on the x-axis.

**Graph**

## Explanation of graph

Since it is a constant term, it is not changing with time. The graph is a straight line, parallel to the x-axis and slope m=0.

### Unit of rate constant

Unit of rate constant for 3rd order reaction is** M ^{-2} sec^{-1}.**

### Half-life

### Half-life is a time in which half of the reactants are converted into products.

Half-life for 3rd order reaction is

**t _{1/2}=**

**Concepts Berg**

- How do you plot an integrated rate law?
- The rate equation for each order of reaction is integrated to relate concentration and time.
- Then plot a graph between concentration on y-axis and time on x-axis.

For zero order reaction, plot a graph between _{t }on the y-axis and time on the x-axis.

For first order reaction, plot a graph between ln** _{t }**on the y-axis and time on the x-axis.

For second order reaction, plot a graph between on the y-axis and time on the x-axis.

For third order reaction, plot a graph between on the y-axis and time on the x-axis.

2. What does the integrated rate law tell you?

integrated rate laws are derived from differential rate laws.

For any chemical reaction, differential rate laws can be integrated with respect to time to give an equation that relates the concentration of reactants or products with the time of reaction.

3. What is the integrated rate equation?

The rate law is a differential equation that relates change in concentration of reactants or products with time.

Differential equations can be integrated to obtain an equation that relates the concentration of reactants or product with time directly which is called integrated rate equation.

4. What is the difference between rate law and integrated rate law?

**Rate law **relates the rate of reaction to the concentration of reaction raised to some power.

For example, rate law for zero order reaction is

=-k_{o}^{o}

**Integrated rate law** relates concentration of species to the time interval.

Integrated rate law for zero order reaction is

_{t}=-k_{o}t+_{o}

5. What is the second-order integrated rate law?

The integrated rate equation for second-order reaction with the same concentration of reactants

=+k_{2}t

Where,

_{t}=conc of reactants at time=t sec

_{o}=conc of reactants at t=0sec

k_{2}=rate constant for 2nd order reaction

6. What is the first-order integrated rate law?

Integrated rate equations for the 1st-order reaction is

ln_{t}=-k_{1}t+ln_{o}

Where,

_{t}=conc of reactants at time=t sec

_{o}=conc of reactants at t=0sec

k_{1}=rate constant for 1st order reaction

7. How do you write a rate law?

Rate law relates the rate of reaction with concentration of reactants raised to some power.

It can be written as

Rate =k

k is a constant depend on order of reaction

‘a’ is some power.

8. How do you find the rate constant k from a graph?

First compare the given integrated rate law equation with the equation of straight line. For example, Integrated rate equations for the 1st-order reaction is

ln_{t}=-k_{1}t+ln_{o}

Compare this equation with y=mx+c then plot a graph between ln_{t }on y-axis and time on x-axis. From the slope of the graph we can calculate the value of the rate constant.

9.Rate vs concentration graph for zero order reaction?

For zero order reaction

_{t}=-k_{o}t+_{o}

Compare eq with the straight-line equation **y=mx+c** and plot a graph between _{t }on the y-axis and time on the x-axis.

**Graph**

This graph shows that the concentration of reactants is decreasing with time. So the graph is a straight line with a negative slope -k_{o} and intercepts _{o}.

10. What is the unit of the pseudo first-order reaction?

11. What is the unit of a rate constant for an nth order reaction?

Unit of rate constant for nth order reaction is

Unit of k=

12. How might differential equations be useful?

Differential equations are playing a very important role in every field of science like biology, chemistry, mathematics, economics and engineering.

In biology, it is used to model different biological processes.

In chemistry, it plays an important role in chemical kinetics. They show the rate of change of one variable with respect to the second variable.

13. Why can’t the base of the exponential function be less than zero?

The base of exponential function must be positive for all real values. If the base is negative, exponential function will be complex function.

**What is the unit for the first-order reaction?**

Unit of rate constant for 1st order reaction is sec^{-1}.