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Class 10 Class 12
For the reaction
: 2A + B + C → A
2B
the rate = k[A] [B]
2 with k = 2.0 x 10–6 mol–2L2s–1.
Calculate the initial rate of the reaction when [A] = 0.1 mol L-1, [B] = 0.2 mol L-1and [C] = 0.8 M. Calculate the rate of reaction after [A] is reduced to 0.06 mol L
–1.

In the reaction
2A + B + C → A2B + C,
there is no change in C, therefore its conc. does not affect the rate of the reaction.
Initial rate = k[A] [B]2
But [A] = 0.1 M,
[B] = 0.2 M
and k = 2 x 10–6 M–2 s–1

Therefore initial rate

Rate= [k] x [A] x [B]2

= 2 x 10–6 M–2 s–1 x (0.1 M) (0.2 mol M)2 = 8 x 10–9 ms–1

From the equation:

2A + B + C → A2B + C,

it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M (due to its 0.04 M has been reacted to 0.02 of B). Thus,

Conc. of A left = [A] = 0.06 M
Conc. of B left = [B] = [0.02 M – 0.02 M]
= 0.018 M

Rate = k[A] [B]2
= 2 x 10–6 M2 S–1 x (0.06 M) (0.18 M)
= 3.89 x 10–9 Ms–1.

263 Views . 5 Shares

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 . 6 Shares

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 . 4 Shares

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 . 3 Shares

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 . 6 Shares

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

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