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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
The rates of a reaction starting with initial concentration 2 x 10–3 M and 1 x 10–3M are equal to 2.40 x 10–4 M s–1 and 0.60 x 10–4 M s–1 respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.

r0 = kR0n(r0)1(r0)2/ = kR01R02n              a  = logr01/r02logR01/R02                   = log(2.40 × 10-4/0.60 × 10-4log (2 × 10-3/1×10-3)                   = log 4log 2=2

Thus reaction is of second order. The rate constant

k=rateA2   = 2.4 ×10-4 Ms-1/(2 × 10-3 M)2   = 0.6 × 102 mol-1 Ls-1.

Use of integrated Rate Equations:
This method is also known as the method of trial and error. The kinetic data are fitted to different integrated rate equations. Whenever if the data fits with the equation for the correct order of the reaction, it will give constant value of rate constant for all data points (concentrations at different times). These equations also lead to straight lines when appropriate function of the concentration is plotted against time ‘t’. For example, for zero reaction, a plot between concentration and time gives a straight line with slope of the line equal to k. Similarly, for the first order reaction, a graph between ln (R) against t gives a straight line with slope equal to – k.
208 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

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

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