The reaction:2NO + Br2 → 2NOBris supposed to follow the following mechanism.(i) NO+Br2 ⇌ fast NOBr2(ii) NOBr2+NO→slow 2NOBrSuggest the rate law expression. from Chemistry Chemical Kinetics Class 12 Nagaland Board
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The reaction:
2NO + Br2 → 2NOBr
is supposed to follow the following mechanism.
(i) NO+Br2  fast NOBr2
(ii) NOBr2+NOslow 2NOBr
Suggest the rate law expression.

The rate expression is derived by step II of the mechanism, as it is the slower one

rate = k[NOBr2][NO] ...(i)

However, NOBr2 is an intermediate and thus its concentration should be replaced from equation (i)
From step (i),

Equilibrium constant, Kc = NOBr2NO2 Br2

          NOBr2 = kc [NO] Br2                  ...(ii)

Then by equation (i) and (ii)

              rate = k' - NO2 Br2

Order of the reaction is 2+1 =3


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.

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.


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.


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.

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.