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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 rate of decomposition of N2O5 in CCl4 solution has been studied at 318 K and the following results have been obtained:

t (in mitt)

0

135

339

683

1680

C (mol L–1)

2.08

1.91

1.68

1.35

0.57

Find the order of reaction and calculate its rate constant.


It can be seen that the rate of reaction does not remain constant with time and, therefore it is not a zero order reaction.

(i) at t = 135 min;            k = 2.303135 minlog2.08191=6.32 × 10-4 min-1(ii) at t = 339 min;           k = 2.303339 minlog2.081.68 = 6.30 × 10-4 min-1(iii) at t = 683 min;          k = 2.303683 minlog2.081.35 = 6.32 × 10-4 min-1(iv) at t = 1680 min;       k = 2.3031680 minlog 2.080.57 = 6.31 × 10-4 min-1

It is observed that the value of k is almost constant for all experimental results and hence it is first order reaction with k = 6.31 x 10
–4 min–1.
166 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

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

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

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