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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:
$\mathrm{A}+2\mathrm{B}⇋2\mathrm{C}+\mathrm{D}$
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)

(iii)

(iv)           $\mathrm{k}=\frac{2.303}{\mathrm{t}}\mathrm{log}\frac{\left[{\mathrm{R}}_{0}\right]}{\left[\mathrm{R}\right]}$

Where

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)

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

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

(i) Order of reaction  = 2 + 1 = 3

(ii) Rate law;  $\mathrm{r}=\mathrm{k}{\left[\mathrm{A}\right]}^{2}\left[\mathrm{y}\right]$

(iii) The rate constant

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

308 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 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 =

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 = $-\frac{1}{2}\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

2088 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
$\mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}$
or

1299 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

(i) Average rate

(ii) Average rate

1717 Views