Following reaction takes place in one step:2NO(g) + O2(g) ⇋2NO2(g)How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one third of its original volume? Will there be any change in the order of the reaction with the reduced volume ? from Chemistry Chemical Kinetics Class 12 Nagaland Board
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Class 10 Class 12
Following reaction takes place in one step:

2NO(g) + O2(g) 2NO2(g)

How will the rate of the above reaction change if the volume of the reaction vessel is diminished to one third of its original volume? Will there be any change in the order of the reaction with the reduced volume ?

The below reaction take place in one step, thus rate of reaction is 

2NO(g) + O2(g)  2NO2(g)

       Rate = kNO2 O2

Suppose x moles of NO and Y moles of O2 are taken in the vessle of volume V litre, then

r1 = kxV2 yV

If the volume of vessel is reduced to V/3, then for the same moles of NO2 and O2

           r2 = kxV/32 yV/32    = 27kyV2 yV


         r2r1 = 27   or   r2 = 27 r1


The rate of the reaction becomes 27 times the initial rate. Order of reaction remains unaffected.
1416 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

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

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

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