﻿ The following results have been obtained during the kinetic studies of the reaction:2A + B → C + D exp.   [A]/ mol L–1 [B]/M Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 x 10–3 II 0.3 0.2 7.2 x 10–2 III 0.3 0.4 2.88 x 10–1 IV 0.4 0.1 2.40 x 10–2 Determine the rate law and the rate constant for the reaction. from Chemistry Chemical Kinetics Class 12 Nagaland Board

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The following results have been obtained during the kinetic studies of the reaction:

2A + B → C + D

 exp. [A]/ mol L–1 [B]/M Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 6.0 x 10–3 II 0.3 0.2 7.2 x 10–2 III 0.3 0.4 2.88 x 10–1 IV 0.4 0.1 2.40 x 10–2

Determine the rate law and the rate constant for the reaction.

Let rate law of reaction be,

where x and y are order of reaction w.r.t. A and B respectively.

From experiments I and IV, we can write
6.0 x 10–3 = k[0.1][0.1]y ...(i)
2.4 x 10–2 = k[0.4][0.1]y ...(ii)

Dividing (ii) by (i), we can write

or

From experiments I and III, we can write

Dividing (iv) by (iii), we can write

or                        $4={2}^{\mathrm{y}}$
or                        $\mathrm{y}=2.$
$\therefore$

Order of reaction w.r.t. A = 1
Order of reaction w.r.t. B = 2
Overall order of reaction = 1 + 2 = 3.
Substituting the values of initial rate of formation of experiment (I) (equ. V) we can write 6.0 x 103 mol L–1 min–1.
= k[0.1 mol L–1] [0.1 mol L–1]2

or

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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

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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

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

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 =

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