For a certain chemical reaction variation in the concentration in [R] vs time(s) plot is given below:For the reaction write/draw(i) What is the order of the reaction?(ii) What are the units of rate constant K? (iii) Give the relationship between K and t1/2 (half-life period).(iv) What does the slope of the above line indicate?(v) Draw the plot Vs. time(s).OrFor a certain chemical reaction:A+2B⇋2C+DThe experimentally obtained information is tabulated below: The experimentally obtained information is tabulated below : Exp. [A]0 [B]0 Initial rate of reaction 1 2 3 4 0.30 0.60 0.30 0.60 0.30 0.30 0.60 0.60 0.096 0.384 0.192 0.768 For this reaction:(i) Derive the order of reaction w.r.t. both the reactants A and B.(ii) Write the rate law.(iii) Calculate the value of rate constant K.(iv) Write the expression for the rate of reaction in terms of A and C.  from Chemistry Chemical Kinetics Class 12 Nagaland Board
Zigya Logo

Chapter Chosen

Chemical Kinetics

Book Chosen

Chemistry I

Subject Chosen

Chemistry

Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

For a certain chemical reaction variation in the concentration in [R] vs time(s) plot is given below:
For the reaction write/draw
(i) What is the order of the reaction?
(ii) What are the units of rate constant K?

(iii) Give the relationship between K and t1/2 (half-life period).
(iv) What does the slope of the above line indicate?
(v) Draw the plot Vs. time(s).
Or
For a certain chemical reaction:
A+2B2C+D
The experimentally obtained information is tabulated below:

The experimentally obtained information is tabulated below :

Exp.

[A]0

[B]0

Initial rate of reaction

1

2

3

4

0.30

0.60

0.30

0.60

0.30

0.30

0.60

0.60

0.096

0.384

0.192

0.768

For this reaction:
(i) Derive the order of reaction w.r.t. both the reactants A and B.
(ii) Write the rate law.
(iii) Calculate the value of rate constant K.
(iv) Write the expression for the rate of reaction in terms of A and C. 


(i) First order

(ii) s-1 or min-1

(iii)           t1/2 = 0.693K

(iv)           k=2.303tlogR0R

Where    
           R0 = Initial conc. of the reactant at t = 0R = Concentration of the reactant at time it.              OrLet the law be r = kAxBywhere              x = order with respect to A                        y = order with respect to B

Taking data from experiment (i) to experiment

(iii), we get
0.096 = k(0.30)x (0.30)y ...(ii)
0.384 = k(0.60)x (0.30)y ...(iii)
0.192 = k(0.30)x (0.60)y ...(iv)
0.768 = k(0.60)x (0.60)y ...(v)
Divide eq. (v) by eq. (iv)

0.7080.192 = 2x = 4 = 22               x=2

Divide eq. (iv) by eq. (ii)

0.1920.096 = 2y = 2'             y=1

Substituting the value of x and y in equation (i), we get rate law


              r = kA2 y

(i) Order of reaction  = 2 + 1 = 3

(ii) Rate law;  r=kA2y

(iii) The rate constant K = rA2 B

Substituting the value of r,  A and B from experiment no. (i)

                     k=0.096(0.30)2(0.30) = 0.0960.027

                        = 3.55 mol-2L-2S-1.

(iv) -dCdt=kB2 A rate of reaction in terms of formation of C   -dAdt = kB2A rate of reaction in terms of formation of A

308 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

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

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