## Grade 12 Chemistry Note

# Chemical Kinetics

**Concept of chemical reaction **

The rate of reaction is defined as decrease in concentration of reactant or increase in product per unit time. It is also defined as the increase of concentration of the product per second.

E.g: A à B (general reaction)

So, Rate = $\frac{{ - \Delta \left[ {\rm{A}} \right]}}{{\Delta {\rm{t}}}}$

**Average rate of reaction **

It is the ratio of reaction measured for long interval of time. Let us consider a general reaction.

A à B

Let, the initial concentration of reactant at time t_{1 }, be [A_{1}] and final concentration be [A_{2}]

So, Average rate (R) = $\left( {{\rm{change\: in\: concentration}}} \right){\rm{\: }}/\left( {{\rm{Change\: in\: time}}} \right)$

Or, -$\frac{{{{\rm{A}}_2} - {{\rm{A}}_1}}}{{{{\rm{t}}_2} - {{\rm{t}}_1}}}$

= $\frac{{ - \Delta {\rm{A}}}}{{\Delta {\rm{t}}}}$

Here, [A_{1}] is greater than [A_{2}]. So, $\Delta $A = (A_{2} – A_{1}). Now, the change in concentration of product is positive.

i.e. C_{2} – C_{1} = +ve.

So, Rate(R) = $\frac{{\Delta {\rm{C}}}}{{\Delta {\rm{t}}}}$

Where, $\Delta {\rm{C}}$ = C_{2} – C_{1}

And $\Delta $t = t_{2} – t_{1}.

** Instantaneous Rate of reaction:**

The average rate obtained by finding the slope of the curve is not always useful. They cause a large interval of time during which the rate of reaction changes significantly. If interval be small the rate is referred to as instantaneous rate and is written as:

Rate = $\frac{{ - {\rm{d}}\left[ {\rm{r}} \right]}}{{{\rm{dr}}}}{\rm{\: }}$

**Factors affecting rate of reaction**

**(i) Concentration:**

Rate depends on concentration of reaction. Initially reaction occurs due to high concentration of reactant. With the passing of time concentration of reactant goes on decreasing which results decrease in rate. It is given in the fig:

Factor affecting rate of reaction by concentration:

**(ii) Temperature:**

According to kinetic molecular theory, average kinetic energy is directly proportional to absolute temperature. The rate of reaction increases with increase in temperature. It is because average kinetic energy of reactant molecule increases as the temperature is increased. For every 10°C rise as the temperature rate is increased by 2 – 3 folds. The effect of temperature in rate can be explained on the basis of Maxwell distribution of molecular velocity.

Fraction of molecules having very high velocity increases with increase in temperature as shown in fig. This result increases K.E. Thus, rate increases fraction with the molecule increase in temperature.

**(iii) Catalyst: **

It is a substance which accelerates or retards the rate of reaction. In general the catalyst increases the rate of reaction. It increases the rate by providing a new path and lowering the activation energy. The effect of catalyst is shown in fig.

Factor affecting rate of reaction by catalyst:

**(iv)Surface area:** The larger the surface area of the reaction the faster is the rate of reaction. It has been observed that if one of the reactants in a solid, then the rate of the reaction depends upon the state of sub – division of the solid. This fact is supported by the general observation that freely divided solids reacts faster than massive substance

**Rate law,Rate constant and its units**

**Rate law**: The rate law of a given reaction depends on the concentration of reactants at a fixed temperature is known as rate law.

Let us consider a general reaction

aA + bB + cC à Products.

So, Rate $ \propto $ [A]^{P}[B]^{a}[C]^{r}

Rate = k[A]^{P}[B]^{a}[C]^{r} ….(i)

Rate constant: The rate of the reaction when the concentration of each of the reacting species is unity at a given temperature is known as Rate constant

[A]= [B]= 1 MOL L^{-1 }then,

R= K [1] [1] or rate,= [ R=K]

**Types of order of reaction**

**First order of reaction:** A chemical reaction in which the rate of reaction depends upon one concentration is known as first order reaction.

Units of first order reaction = min^{-1} or sec^{-1}.

**Second order reaction:** A chemical reaction in which the rate depends upon second power of initial concentration of reactant is called second order reaction.

Unit of second order reaction:= mol^{-1} L sec^{-1}.

** Third order reaction: **A Chemical reaction in which rate depends upon third power of initial concentration of the reactant is called first order reaction. Unit of third order reaction = mol^{-2} L^{2} sec^{-1}.

**Zero order reaction: **A chemical reaction in which rate of reaction is independent of concentration of the reactant is called third order reaction.

Unit of zero order reaction: mol^{-1 }L^{-1 }S^{-1 }

**Order and molecularity of reaction**

The difference between the order and molecularity of reaction are given below:

Order of reaction |
Molecularity of reaction |

It is the total number of concentration variables which affect the rate of reaction. |
It is the total no. of chemical species involved in the reaction. |

It is experimentally determined value. |
It is theoritically concept. i.e. assigned parameters. |

It changes with change in temperature and pressure. |
It doesn’t change with change in temperature and pressure. |

**Integrated law for first order reaction**

Let us consider general first order reaction,

A à Product.

Let the initial concentration of A is ‘a’ moles and x moles L^{-1} of it gets changed to products after ‘t’. Thus, the concentration of reactant A left after time ‘t’ is (a – x) moles per liter i.e.

A à product

A = ‘a’ moles L^{-1}

(a – x) moles L^{-1}

Products: Initially (At time ‘t’)

According to law of mass action the rate of a first reaction of any time ‘t’ be directly proportional to the concentration of reactant present in the system of that instant. Therefore, it is given by:

Rate = $\frac{{ - {\rm{d}}\left[ {\rm{A}} \right]}}{{{\rm{dt}}}} = \frac{{ - {\rm{d}}\left( {{\rm{a}} - {\rm{x}}} \right)}}{{{\rm{dt}}}} = {{\rm{k}}_1}\left( {{\rm{a}} - {\rm{x}}} \right)$

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = {{\rm{k}}_1}\left( {{\rm{a}} - {\rm{x}}} \right)$….(i)

Rearranging equation (i), we get,

Or, $\frac{{{\rm{dx}}}}{{{\rm{a}} - {\rm{x}}}} = {{\rm{k}}_1}{\rm{dt}}$

On integrating above equation, we get,

Or, $\mathop \smallint \nolimits^ \frac{{{\rm{dx}}}}{{{\rm{a}} - {\rm{x}}}} = \mathop \smallint \nolimits^ {{\rm{k}}_1}{\rm{dt}}$

= -ln(a – x) = k_{1}t + I …(ii)

Where, I is integration constant, When t = 0, x = 0,

Thus, -ln(a – 0) = k_{1} * 0 + I

So, I = -lna.

Substituting, the value of I in equation (ii), we have,

Or, - ln(a – x) = k_{1}t – lna

Or, k_{1} = $\frac{1}{{\rm{t}}}$.ln$\frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$

Or, K_{1} = $\frac{{2.303}}{{\rm{t}}}.\log \frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$ ….(iii)

So, Equation (iii) is known as integrated rate law for first order reaction.

The value of rate constant of a first order reaction is independent of units of concentration used. This can be proved as follows:

Let the unit of concentration be changed. So, that in new unit system the value of concentration becomes ‘m’ times of its original value, then from the equation (iii), we can write,

Or, K_{1} = $\frac{{2.303}}{{\rm{t}}}.\log \frac{{{\rm{ma}}}}{{{\rm{ma}} - {\rm{mn}}}}$

Or, K = $\frac{{2.303}}{{\rm{t}}}.\log \frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$

This is same as equation (iii).

**Half life reaction of first order**

The time required to decompose exactly the half of the initial concentration of reactant is known as half life period. It is denoted by t_{1/2}.

For 1^{st} order reaction we can write,

K_{1} = $\frac{{2.303}}{{\rm{t}}}$.log $\frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$, where initial concentration of reactant and (a – x) concentration after time ‘t’ seconds.

For t_{1/2} life period,

t = t_{1/2} and (a – x) = $\frac{{\rm{a}}}{2}$.

So, K_{1} = $\frac{{2.303}}{{{{\rm{t}}_{\frac{1}{2}}}}}$.log$\left( {\frac{{\rm{a}}}{{\frac{{\rm{a}}}{2}}}} \right)$

Or, t_{1/2} = $\frac{{2.303}}{{{{\rm{K}}_1}}}.{\rm{log}}2$

Or, t_{1/2} = $\frac{{0.693}}{{{{\rm{K}}_1}}}$

i.e. t_{1/2}$ \propto $$\frac{1}{{{{\rm{k}}_1}}}$.

Thus, t_{1/2}of the first order reaction is independent of the initial concentration of reactant.

Or,

In a first order reaction,

A = product.

t = 0 [A] = a product = 0

t, [A] = (a – x) , product = x.

Then, Rate = $\frac{{ - {\rm{d}}\left[ {\rm{A}} \right]}}{{{\rm{dt}}}} = \frac{{{\rm{dx}}}}{{{\rm{dt}}}} = {\rm{k}}\left[ {\rm{A}} \right]$

Or, $\frac{{{\rm{dx}}}}{{{\rm{dt}}}} = {\rm{k}}\left( {{\rm{a}} - {\rm{x}}} \right).$

Or, $\frac{{{\rm{dx}}}}{{{\rm{a}} - {\rm{x}}}} = {\rm{dt}}.{\rm{K}}$

Integrating on both sides,

Or, ln(a – x) = kt + c

At t = 0, x = 0

Then –lna = c

Putting the value of C,

- ln(a – x) = kt – lna

Or, kt = ln $\frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$

Or, t = $\frac{1}{{\rm{k}}}.{\rm{ln}}.\frac{{\rm{a}}}{{{\rm{a}} - {\rm{x}}}}$

At, t_{1/2}, X = $\frac{{\rm{a}}}{2}$.

Then, t_{1/2} = $\frac{1}{{\rm{k}}}{\rm{ln}}.\frac{{\rm{a}}}{{{\rm{a}} - \frac{{\rm{a}}}{2}}}$

Or, $\frac{{{{\rm{t}}_1}}}{2} = \frac{1}{{\rm{k}}}{\rm{ln}}2$

Or, $\frac{{{{\rm{t}}_1}}}{2} = \frac{{0.693}}{{\rm{K}}}$

[So, $\frac{{{{\rm{t}}_1}}}{2} = \frac{{0.693}}{{\rm{k}}}$]

Thus, half – life of first order is independent of initial concentration

**Collision theory of reaction rate:**

According to this theory all molecules collision are not effective only a fraction of molecules colliding with sufficient kinetic energy leads to the formation of product called effective collision.

Thus, molecules collision occurring with sufficient kinetic energy which leads to the formation of product is called effective collision. The molecules collision occurring with insufficient K.E. and improper orientation is called inefficient collision.

Consider the formation of hydrogen iodide by the reaction between H_{2} and I_{2}.

H_{2} + I_{2} = 2HI

This can be diagrammatically represented as:

Every substance has definite amount of energy due to “to and fro” motion is called average energy [E_{av}]. In order to cause a chemical change, reacting species should cross over energy barrier as shown in energy profile diagram. The minimum amount of energy required to cross the energy barrier is called activation energy [E_{a}]. The minimum amount of energy associated with activated complex that can cause a chemical change is called threshold energy (E_{th}). Therefore, threshold energy is the sum of average (E_{a}) and activation energy (E_{av}).

So, E_{th} = E_{av} + E_{a}

So, E_{a} = E_{th} - E_{av}

Thus, activation energy is the additional amount of energy supplied to reactant to cause a chemical change.

This theory is called collision theory for bio – molecular reaction

Collision theory of reaction rate

**Concept of activation energy as the energy barrier, activate complex and effect of catalyst on the rate of reaction **

Every substance has definite amount of energy due to “to and fro” motion is called average energy [E_{av}]. In order to cause a chemical change, reacting species should cross over energy barrier as shown in energy profile diagram. The minimum amount of energy required to cross the energy barrier is called activation energy [E_{a}]. The minimum amount of energy associated with activated complex that can cause a chemical change is called threshold energy (E_{th}). Therefore, threshold energy is the sum of average (E_{a}) and activation energy (E_{av}).

So, E_{th} = E_{av} + E_{a}

So, E_{a} = E_{th} - E_{av}

Thus, activation energy is the additional amount of energy supplied to reactant to cause a chemical change.

This theory is called collision theory for bio – molecular reaction

**Effect of catalyst on rate of reaction **

It is a substance which accelerates or retards the rate of reaction. In general the catalyst increases the rate of reaction. It increases the rate by providing a new path and lowering the activation energy. The effect of catalystis in shown in fig.

Factor affecting rate of reaction by catalyst