Rate of Chemical Reactions
Chemical reaction involves transformation of reactant into product.Rate of chemical reaction is the measure of how fast these changes are taking place.Some reactions occur very rapidly;others very, very slowly.For example, ionic reactions are very fast, while those taking place in water treatment plant may last up to few days.
Contents
Chemical reaction is a thing of probability
To get the rough idea what actually happens during chemical reaction, think about molecules of different reactants. Molecules are travelling every time.Molecules of solid are least mobile while gaseous molecules have high freedom for mobilization. According to collision theory, for reaction to take place, reacting molecules must collide. If this was the only case, rate of reaction would be too fast because there are lot of molecule colliding with each other every second. But every collision is not fruitful to yield a new product. For this, reacting molecules must have sufficient energy to break old bonds and form new bonds.If colliding molecule are not having this minimum energy,nothing much happen and they bounce back. This is not only thing, they must also collide with proper orientation . These two requirements make these collision less fruitful.
This picture also helps us to understand effect of temperature on rate of reaction. With increase in temperature, energy of each molecule also increases in the form of their kinetic energy.Now, there are more molecule which can cross that minimum energy barrier which explain why rate increases with increase in temperature.
Average rate of reaction
Average rate of chemical reaction is defined as the change in number of moles of reactants and products in unit time for unit volume of reaction mixture.
\[\text{Average rate of reaction} =\frac{\text{change in no. mole}}{\text{(volume)(time)}}\]
For reactions in which volume of reaction mixture remains constant throughout reaction (e.g. Liquid and solid phase reaction) ,
\[\text{Average rate of reaction} =\frac{\text{change in concentration}}{\text{time}}\]
As concentration of a substance is usually expressed as number of mole of a substance in unit volume of mixture, unit of rate of reaction turns out to be \(mol/cm^3.s\) .
\[A+2B \to 3C+2D\]Initially, concentration of reactants A and B is 6 \(mol/cm^3\).What will be average rate of reaction when concentration of A drops to 4 \(mol/cm^3\) in 2 second?
During the above change, concentration of A decrease from 6 \(mol/cm^3\) to 4 \(mol/cm^3\) in 2 second.Hence by definition, rate of reaction is given by
\[=\frac{-\Delta C}{\Delta t}=\frac{-(4-6)}{2}=\boxed{1\space mol/cm^3.s}\]
For reactants, concentration decreases as time increases .Hence change in concentration of reactant molecules will be negative(.e. final minus initial), from which rate of reaction turns out to be negative. This negative sign is useless since rate of reaction is the measure how fast conversion of reactant into product is happening.To avoid this,a extra negative sign is put before change in concentration of reactants in definition of reaction rate.
What is the rate of reaction with respect reactant B?
In above reaction,1 mole of A reacts with 2 moles of B. Two moles of A that reacted will require 4 moles of B.Hence, during the same interval, concentration of B changes from 6 \(mol/cm^3\) to 2 \(mol/cm^3\).Therefore, Rate of reaction with respect to reactant B is
\[=\frac{6-2}{2}=\boxed{2 \space mol/cm^3}\]
Similarly is the case of product also. Rate of reactions with respect to different reactants and products are related as
\[ \frac{-r_a}{a}=\frac{-r_b}{b}=\frac{r_c}{c}=\frac{r_d}{d} \] For above reaction,various reaction rate are related as
\[ \frac{-r_a}{1}=\frac{-r_b}{2}=\frac{r_c}{3}=\frac{r_d}{2} \]
Instantaneous rate of reaction
Average rate of reaction gives lumped or rough estimate of rate.To determine rate of reaction at each instant, we need to define Instantaneous rate of reaction.
Rate of chemical reaction at particular instant is defined as the infinitesimal change in number of mole of reactants and products in infinitesimal time for unit volume of reaction mixture
\[r =\frac{dN}{Vdt}\]
For reaction other than gaseous phase,above definition becomes
\[r =\frac{dN}{Vdt}=\frac{dN/V}{dt}=\frac{dC}{dt}\]
You must have noticed why \(r_a=\frac{dC}{dt}\) only for liquid and solid phase reaction.We know solid and liquid are in-compressible.Their volume do not appreciably change during reaction hence V (Volume of reaction mixture) remains constant throughout reaction and can be taken inside differential sign which finally leads to \(r_a=\frac{dC}{dt}\).
For above reaction, concentration of reactant A changes with time according to following relation.\[C_a=6e^{-t}\]
Find rate of reaction w.r.to A at 2 second.
This time,we do not have to evaluate average reaction rate.Instead We have to find the rate at the moment when two second has elapsed.This can be found easily by differentiating concentration w.r. to time and putting t=2.
\[r_a=\frac{-dC_a}{dt}=\frac{-d(6e^{-t})}{dt}=6\times (e^{-t})\]
Now putting t=2 yields
\[r_a=\boxed{0.81\space mol/cm^3.s}\]
Rate Law
Rate of reaction is directly proportional to the concentration of reactants with each concentration terms raised to a experimentally determined exponents.
\[-r_a\propto C_a^rC_b^s\]
\[-r_a=k C_a^rC_b^s\]
k in above equation is called rate constant.It is also called specific rate law as rate of reaction equals rate constant when concentration of all reactant is unity.
r and s in above equation are called order of reaction with respect to reactant A and B respectively.Overall order of reaction is the sum of all individual orders which is r+s in this case.Order of reaction may be whole number fraction and even negative in some rare case.
When concentration of reactant in reaction
\[A\rightarrow B\] is increased by 5 times,the rate increase 25 times. What will be the order of a reaction ?
From rate law, we have
\[-r_a=k C_a^r\]
This problem is all about finding value of r.For this Consider the ratio of rate at two different concentrations.
\[\frac{r_{a2}}{r_{a1}}=(\frac{C_2}{C_1})^r\]
We have from problem,\(\frac{r_2}{r_1}=25\) and \(\frac{C_2}{C_1}=5\) which gives,
\[25=5^r\] Hence order of reaction is
\[r=2\]
Half life time
Half life time:Time required for concentration of reactant to become half of its initial value.It is denoted by \(t_{1/2}\).
Consider a general \(n^{th}\) order reaction whose rate expression is given by
\[-r_a=kC^{n}\]
For constant volume system, we have \[-r_a=\frac{-dC}{dt}=kC^{n}\]
Solving this differential equation yields integrated rate expression as
\[-\frac{c^{-n+1}}{-n+1}=kt+N\]
where N is a constant of integration whose value can be found as follow:At t=0,C=\(C_0\)
Hence,
\[N=\frac{1}{(n-1)c^{n-1}}\]
Putting this value of N in above equation gives
\[k(n-1)t=(\frac{1}{C^{n-1}}-\frac{1}{{C_0}^{n-1}})\]
For half life period, put \(t=t_{1/2}\) and \(C=\frac{C_0}{2}\).Solving we get
\[t_{1/2}=\frac{2^{n-1}-1}{(n-1)kC^{n-1}}\]
To find out half life time for first order reaction, put n=1 in above equation
\[t_{1/2}=\frac{ln(2)}{k}\]
We see that half life time for first order reaction is independent of initial concentration of reactant.
Factors that affect rate law
- Temperature
Rate of reaction increases with increase in temperature.Temperature dependency of reaction rate is given by the general expression
\[k=k_0T^me^{-E_a/RT} \quad and \quad 0\leq m \leq 1 \] where
\(E_a\) =Activation energy
T=temperature
R=universal gas constant
Most widely used theory about temperature dependency is Arrhenius theory (m=0) which can be explained as to reaction to take place,reacting molecules must possess minimum energy.Number of molecules having this minimum energy is found to be proportional to \(e^{-E_a/RT}\).Other theories are Collision theory (m=0.5) and Transition state theory(m=1).
Prove that for most of the reaction at room temperature, rate of reaction gets doubled on 10 K rise in temperature.
We have,
\[k=k_0e^{-E_a/RT} \] Taking ratio of rate constant at two different temperature,
\[\frac{k_2}{k_1}=e^{-\frac{-E_a}{R}(\frac{1}{T_2}-\frac{1}{T_1})}\]
Now putting, \(T_1\)=298K , \(T_2\)=308K and \(E_a=53\) kJ/mol
\[\boxed{\frac{k_2}{k_1}=2}\]
- Pressure
Pressure affects the rate of reaction, especially when you look at gases. When you increase the pressure, the molecules have less space in which they can move. That greater density of molecules increases the number of collisions. When you decrease the pressure, molecules don't hit each other as often and the rate of reaction decreases.