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Chemical Kinetics

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Chemistry I

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Chemistry

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12

(i) Define specific reaction rate.
(ii) Define Half-life period of a chemical reaction. Also obtain the expression for half-life period.

Or

Derive the general for of the expression for the half-life of a first order reaction.


(i) Specific reaction rate. This is also called rate constant or velocity constant. Specific reaction rate may be defined as the rate of reaction under specific conditions, when the product of concentration of the reactants is unity.

(ii) Half-life period of a chemical reaction in which the concentration of reactant is reduced to half of the intial value of concentration.
It is denoted by tl/2 or t0.5.

The rate equation for a reaction of first order is expressed as:
                 k= 2.303tlogaa-x


or               t=2.303klogaa-x


   When      t=12,  x = 0.5 a or a2

   On substituting these values, we get

          t1/2 2.303klogaa-0.5a=2.303klog2.or      t1/2 = 0.693k                   ...(i)

The relation (i) is the required expression for half-life period of first order reaction.

516 Views

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

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

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

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=2.303klogR0R
or           
                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

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]
1/2[B]2

Order of reaction = 12+2 = 2.5.

1504 Views