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Chemical Kinetics

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Chemistry I

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
What is the effect of temperature on the rate constant of reaction? How can this effect of temperature on rate constant be respresented quantitatively?

The rate constant of reaction increases with increase of temperature. This increase is generally two fold to five fold for 10 K rise in temperature. This is explained on the basis of collision theory. The main parts of collision theory are as follows:

(i) For a reaction to occur, there must be collision between the reacting species.

(ii) Only a certain fraction of total collisions are effective in forming the products.

(iii) For effective collisions, the molecule must possess the sufficient energy (equal or greater
than threshold energy) as well as proper orientation.
On the basis above conclusions, the rate of reaction is given by
Rate =f x 2 (where f is the effective collisions and is total number of collisions per unit volume per second).

Quantitatively, the effect of temperature on the rate of a reaction and hence on the rate constant k was proposed by Arrhenius. The equation, called Arrhenius equation is usually written in the form

K=Ae-Ea/RT            ...(i)
where A is a constant called frequency factor (because it gives the frequency of binary collisions of the reacting molecules per second per litre, E
0 is the energy of activation, R is a gas constant and T is the absolute temperature. The factor e–Ea/RT gives the fraction of molecules having energy equal to or greater than the activation energy, Ea.

The energy of activation (Ea) is an important quantity and it is characteristic of the reaction. Using the above equation, its value can be calculated.

Taking logarithm or both sides of equation (i), we get,
In k = In A - EaRT1

If the value of the rate constant at temper-atures T
1 and T2 are k1 and k2 respectively, then we have

 In k1 = In A - EaRT1            ...(ii)In k2 = In A - EaRT2            ...(iii)

Subtracting eqn. (i) from eqn. (ii), we get

   In k2 - In k1 = -EaRT2+EaRT1                       = EaRT1+EaRT2

or                          Ink2k1 = EaR+1T1-1T2                                         = EaRT2-T1T1T2

or       logk2k1 = Ea2.303RT2-T1T1 T2

Thus knowing the values of the constant k
1 and k2 at two different temperature T1 and T2, the value of Ea can be calculated.

108 Views . 11 Shares

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M is 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Given that 
Initial concentration, [R1] = 0.03
Final concentration, [R2] = 0.02
Time taken ∆t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )
The formula of average rate of change 

rav =-Rt =[P]t

(i) Average rate
                     = (0.03 - 0.02) M25 × 60 sec= 0.01 M25×60 s = 6.66 M s-1
(ii) Average rate
                        = (0.03-0.02)M25 min =  0.01 M25= 0.0004 Ms-1.
1717 Views . 4 Shares

In a reaction 2A → Products, the concentration of A decreases from 0.5 mol to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval.

Given that 
Initial concentration [A1] =0.5
Final concentration [A2] =0.4
Time is  = 10 min

Rate of reaction = Rate of disappearance of A.

Rate of reaction = -12[A]t
                         = - 12(0.4-0.5) mol L-110 minute= 0.1 mol L-15 minutes= 0.005 mol litre-1 min-1.
2088 Views . 3 Shares

For a reaction, A + B → Product; the rate law is given by, r = k[ A]1/2 [B]2. What is the order of reaction?

The order of the reaction is sum of the powers on concentration.
So that sum will 

r = k[A]

Order of reaction = 12+2 = 2.5.

1504 Views . 6 Shares

A first order reaction has a rate constant 1.15 x 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g?

Given that 
Initial quantity, [R]o= 5 g
Final quantity, [R] = 3 g 
Rate constant, k = 1.15 x 10−3 s−1
Formula of 1st order reaction,
We know that
                t = 2.3031.15 × 10-3log53   = 2.3031.15 × 10-3(log 5 - log 3)        = 2.3031.15 × 10-3(0.6990 - 0.4771)     = 2.303 × 0.22191.15 × 10-3     = 2.303 × 0.2219 × 10001.15      = 444 sec.
1299 Views . 10 Shares

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Let the reaction is X →Y

This reaction follows second order kinetics.
So that, the rate equation for this reaction will 
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
= ka2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get 
Rate, R2 = k (3a)2
= 9ka2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1 
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.

972 Views . 6 Shares

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