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Rate law is the reaction in which the reaction rate is determined in terms of molar concentration of reactants with all the terms raised to some power, which can or cannot be the same as the stoichiometric coefficient of the reacting things in a specific balanced chemical equation. Consider a rate of reaction, aA+bB → cC+dD where, a, b, c and d are the stoichiometric coefficients of reactants and products. The rate expression for this reaction is, Rate∝AxBy where x and y exponents can or cannot be equal to the stoichiometric coefficients (a and b) of the reactants and products. Also, the above equation can also be written as, Rate=K AxBy, where, k is a rate constant which means it is a proportionality constant.
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Key Terms: Rate law, Proportionality constant, Order of reaction, Molecularity, Differential method, Power, Rate of reaction, Reactants
Rate Law
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Rate law is the representation of rate of reaction in the terms of the molar concentration of the reactants participating in a reaction given to some power. Rate law is also known as the rate expression or the rate equation.
Rate law expression=-d[R]/ dt=k [NO]2 [O2]
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Method of Determining Order of Reaction
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There are some methods to determine the order of reaction:
Graphical Method
Rate Law expressions for many types of reactions,
dx/dt=k zero order
dx/dt=k[a-x] first order
dx/dt=k[a-x]2 second order
dx/dt=k[a-x]n nth order
The plot between dx/dt vs concentration is shown and if it is a straight line passing through origin then corresponding concentration power is the order of reaction.
Graphical Method
Initial Rate Method
The values are given in terms of different rates at different concentrations; so, we calculate the rate expression and then find the order of reaction.
Example:
Reaction A+B → Product
| S.No. | Concentration of A | Concentration of B | Rate |
|---|---|---|---|
| 1 | x1 | y1 | r1 |
| 2 | x2 | y2 | r2 |
| 3 | x3 | y3 | r3 |
According to rate law:
\(\text {Rate_{1}} = k(X_1)^m (Y_1)^n ............(i)\)
\(\text {Rate_{2}} = k(X_2)^m (Y_2)^n ............(ii)\)
\(\text {Rate_{3}} = k(X_3)^m (Y_3)^n ............(iii)\)
Here we have three quantities k, m, n which are unknown and we have three equations so, the solution is possible and k, m, n have a unique value.
So, the order of reaction=m+n.
Integration Methods
In this method the concentration of reactants at some points of time are calculated and put into the rate equation of different order, the rate equation which gives the specific constant value of k is the given equation of that reaction.
Fractional Change Method
Time required for any fractional change is related with the starting concentrations of reactants as:
Isolation Method
In isolation method, one reactant is chosen in large excess and the order is checked with respect to the other process is repeated and the real order is the sum of the order of all the isolated reactions.
Zero order Reaction
Mathematical expression for the zero order reaction:
\(\text{Rate equation} = \frac{dx}{dt}- = k(A) ^{\circ}\)
\(= \frac{dx}{dt} - = k\)
\(\text {On integrating x = kt + c}\)
Rate of reaction is independent of concentration. Where c ≠ O
Unit of Rate Constant: mol. liter-1 time-1
Examples:
First Order Reaction
Mathematical expression for the first order reaction:
Pseudo First Order Reaction
When the order of reaction looks like 2 but one component is in the large excess then it behaves like first order, and this type of reaction is called pseudo first order reaction.
Examples:
when water is taken as a solvent i.e., in excess.
- Rate law: rate = k[H2O2] describes a reaction that is the first order in hydrogen peroxide and the first order overall.
- Rate law: rate = k[C4H6]2 describes a reaction that is the second order in C4H6 and second-order overall.
- Rate law: rate = k[H+] [OH−] describes a reaction that is first order in H+, first-order in OH−, and second-order overall.
Rate Constants for Common Reaction Orders
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Rate constants for common reaction orders are:
| Reaction Order | Units of k |
|---|---|
| (m+n) | mol1−(m+n) L(m+n) −1s−1 |
| zero | mol/L/s |
| first | s−1 |
| second | L/mol/s |
| third | mol−2 L2 s−1 |
*Note- the units in the table can also be shown in terms of molarity (M) instead of mol/L.
Also, the units of time other than the second (such as minutes, hours, days) may be used, depending on the situation.
Things to Remember
- Rate law is also known as the rate expression or the rate equation.
- So, the order of reaction = m+n
- Unit of Rate Constant: mol. litre-1 time-1
- Rate law expression = -d[R]/ dt = k [NO]2 [O2]
- Molecularity is the sum of the number of molecules undergoing the covalency change in a balanced stoichiometric equation of a reaction.
- Basically, zero order, first order, second order and a third order reaction are of much importance.
- Rate Law expressions for many types of reactions,
- dx/dt=k zero order
- dx/dt=k[a-x] first order
- dx/dt=k[a-x]2 second order
- dx/dt=k[a-x] n nth order
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Sample Questions
Ques. A certain rate law is given as Rate=k[H2] [Br2]12 Rate=k[H2] [Br2]12. What is the reaction order? (3 Marks)
Ans. x=1, y=12 x=1, y=12
reaction order = x + y = 1 + 12 = 32 reaction order = x + y = 1 + 12 = 32
The reaction is first-order in hydrogen, one-half-order in bromine, and 3232-order overall.
Ques. The reaction between nitric oxide and ozone, NO(g)+O3(g)→NO2(g)+O2(g)NO(g)+O3(g)→NO2(g)+O2(g), is first order in both nitric oxide and ozone. (2 Marks)
Ans. The rate law equation for this reaction is: Rate=k[NO]1[O3]1 Rate=k[NO]1[O3]1.
The overall order of the reaction is 1 + 1 = 2.
Ques. Writing Rate laws from reaction Orders.
An experiment shows the reaction of nitrogen dioxide with carbon monoxide:
NO2(g) + CO(g) → NO(g) + CO2(g)
is second-order in NO2 and zero-order in CO at 100°C. What is the rate law for the reaction? (3 Marks)
Ans. The reaction will have the form:
rate = k[NO2]m[CO]n
The reaction is second order in NO2; thus m=2. The reaction is zero order in CO; thus n=0. The rate law is:
rate = k[NO2]2[CO]0 = k[NO2]2
Remember that a number raised to the zero power is equal to 1, thus [CO]0=1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO2. When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
Ques. In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel:
CH3OH+CH3CH2OCOCH3 → CH3OCOCH3+CH3CH2OH (2 Marks)
Ans. The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
rate = k[CH3OH]
Ques. What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction? (2 Marks)
Ans. It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates, to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.
Ques. Determine the rate law and the rate constant for the reaction at 25 °C. (2 Marks)
Ans. The rate law will have the form:
rate = k[NO]m[O3] n
We can determine the values of m, n, and k from the experimental data using the following three-part process:
Ques. For the reaction given by 2NO+O2 → 2NO2, the rate equation is:
Rate = k[NO]2[O2]
Find the overall order of the reaction and the units of the rate constant. (2 Marks)
Ans. The overall order of the reaction = sum of exponents of reactants in the rate equation = 2 + 1 = 3
The reaction is a third-order reaction. Units of rate constant for ‘nth’ order reaction = M ( 1 – n). s – 1
Therefore, units of rate constant for the third-order reaction = M (1 – 3). s – 1 = M – 2. s – 1 = L2.mol-2. s-1
Ques. For the first-order reaction given by 2N2O5 → 4NO2 + O2, the initial concentration of N2O5 was 0.1M (at a constant temperature of 300K). After 10 minutes, the concentration of N2O5 was found to be 0.01M. Find the rate constant of this reaction (at 300K). (3 Marks)
Ans. From the integrated rate equation of first-order reactions:
k=(2.303/t) log([R0]/[R])
Given, t=10 mins=600 s
Initial concentration, [R0]=0.1M
Final concentration, [R]=0.01M
Therefore, rate constant, k=(2.303/600s) log (0.1M/0.01M)=0.0038 s-1
The rate constant of this equation is 0.0038 s-1
Ques. Use the provided initial rate data to derive the rate law for the reaction whose equation is:
OCl−(aq) +I−(aq) → OI−(aq)+Cl−(aq)
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction. (2 Marks)
Ans. rate=k [O Cl−]2[I−]10.00184
k (0.0040)2(0.0020)1
k=5.75×104mol−2L2s−1.
Ques. If a reaction is given by aA+bB → cC+dDaA+bB → cC+dD
Where a, b, c, and d denote the stoichiometric coefficients of the reactants and products, the rate equation for the reaction is given by: (3 Marks)
Ans. Rate∝[A]x[B]y Rate∝[A]x[B]y
⇒Rate=k[A]x[B]y ⇒ Rate=k[A]x[B]y
[A][A] & [B][B] denote the concentrations of the reactant side AA & BB.
The proportionality constant ‘k′ ‘k′ is the rate constant for the reaction.
xx & yy denote the partial reaction orders for the reactant side AA & BB (This may or may not be equal to their stoichiometric coefficients aa & bb).
Ques. If the rate constant of a reaction is k=3 × 10-4 s-1, then identify the order of the reaction. [Comptt. All India 2013] (1 Mark)
Ans. S-1 is the unit for rate constant of first order reaction.
Ques. Write the unit of rate constant for a zero-order reaction. [Comptt. All India 2013] (1 Mark)
Ans. Mol L-1 S-1 is unit of rate constant for a zero-order reaction.
Ques. Define rate of reaction. [Comptt. Delhi 2016] (1 Mark)
Ans. The change in concentration of reactant or product per unit time is called rate of reaction.
Ques. A reaction is of second order with respect to a reactant. How will the rate of reaction be affected if the concentration of this reactant is
(i) doubled, (ii) reduced to half? [Delhi 2009] (3 Marks)
Ans. Since Rate=K[A]2
For second-order reaction Let [A]=a then Rate=Ka2
(i) If [A]=2a then Rate=K (2a)2=4 Ka2
∴Rate of reaction becomes 4 times
(ii) If [A]=a2 then Rate=K (a2)2=Ka24
∴ Rate of reaction will be 14 th.
Ques. Define the following:
(i) Order of a reaction
(ii) Activation energy of a reaction [All India 2009] (3 Marks)
Ans. (i) Order of a reaction:
It is the sum of powers of molar concentrations of reacting species in the rate equation of the reaction.
It may be a whole number, zero, fractional, positive or negative.
It is experimentally determined.
It is meant for the reaction and not for its individual steps.
(ii) Activation energy of a reaction: The minimum extra amount of energy absorbed by the reactant molecules to form the activated complex is called activation energy.
Ques. Write two differences between ‘order of reaction’ and ‘molecularity of reaction’. [Delhi 2014] (2 Marks)
Ans.
| Order of reaction | Molecularity of reaction |
|---|---|
| It is the sum of tire concentration terms on which die rate of reaction actually depends. | It is the number of atoms, ions or molecules that must collide with one another simultaneously so as to result into a chemical reaction. |
| It can be fractional as well as zero. | It is always a whole number. |
Ques. (a) A reaction is first order in A and second order in B.
(i) Write a differential rate equation.
(ii) How is the rate affected when the concentration of B is tripled?
(iii) How is the rate affected when the concentration of both A and B is doubled?
(b) What is the molecularity of a reaction? [Comptt. All India 2009] (4 Marks)
Ans.
(a)
(i) Differential rate equation:
dxdt=rate=K [A]1 [B]2
(ii) Rate, r1=K [A]1[B]2 …………… (i)
When concentration of B is increased three times then
Rate, r2=K [A]1 [3B]2 ………..(ii)
Dividing equation (ii) by (i) we get
r2=9r1 rate increases by n*ne times.
(iii) When concentration of both A and B are doubled, then
r3=K [2A]1 [2B]2 ………….. (iii)
Dividing equation {Hi) by (t), we get
r3=8r1
Hence the rate increases by eight times.
(b) The number of reacting species (atoms, ions or molecules) taking part in an elementary reaction is called Molecularity of a reaction.
Ques. (a) The decomposition of A into products has a value of K as 4.5 × 103 s-1 at 10°C and energy of activation 60 kj mol-1. At what temperature would K be 1.5 × 104 s-1?
(b) (i) If half-life period of a first order reaction is x and 3/4,th life period of the same reaction is y, how are x and y related to each other?
(ii) In some cases it is found that a large number of colliding molecules have more energy than threshold energy, yet the reaction is slow. Why? [Comptt. Delhi 2013] (5 Marks)
Ans. (a) Given: K1=4.5 × 103 s-1,
T1=10K+273K=283K
K2=1.5 × 104 s-1, T2=?
Ea=60 KJ mol-1
Using formula :
∴ Temperature, T2 will be=297° – 273°=24° C
(b) (i) t1/2=0.693K (For first order reaction)
t3/4=K ⇒ t3/4=1.3864K
According to condition
(The value 1.3864 is double of 0.693)
From the above equation it is clear that
t3/4=2t1/2 ∴ y=2X
(ii) It is due to improper orientation of the colliding molecules at the time of collision.
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