### 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

(a) The decomposition of N2O5(g) is a first order reaction with a rate constant of 5 x 10–4 sec–1 at 45°C.
i.e., 2N2O5(g) = 4NO2(g) + O(g)
If initial concentration of N2O5 is 0.25, calculate its concentration after two minutes. Also calculate half life for the decomposition of N2O5(g).
(b) For an elementary reaction: 2A + B → 3C The rate of appearance of C at time ‘t’ is 1.3 x 10–4 mol l–1 s–1. Calculate at this time:
(i) Rate of reaction (ii) Rate of disappearance of A.

Rate constant K = 5 x 10–4 sec. Initial concentration [A]0 = 0.25 M Final concentration [A]t =? Time taken by the reaction, t = 2 min.
For a first order reaction, rate constant (K) is given by

(b)

(i) The rate of appearance of C at time t

(ii) Rate of disappearance of A

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 =

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

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

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

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.

Do a good deed today
Refer a friend to Zigya