ï»¿ e reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table: exp.   [A]/ mol L–1 [B]/M Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 2.0 x 10–2 II - 0.2 4.0 x 10–2   III 0.4 0.4 - IV - 0.2 2.0 x 10–2 from Chemistry Chemical Kinetics Class 12 Nagaland Board

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e reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
 exp. [A]/ mol L–1 [B]/M Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 2.0 x 10–2 II - 0.2 4.0 x 10–2 III 0.4 0.4 - IV - 0.2 2.0 x 10–2

Rate law for the reaction is given by:
Rate = k [A]1 [B]0 = k[A]
2.0 x 10–2 mol L–1 min–1 = k[0.1 mol L–1]
Rate constant,

Rate constant = k = 0.2 min–1.
(i) In experiment II Rate = k[A]

(ii) In experiment III

Rate = k[A]
= 0.2 min–1 x 0.4 M
= 0.08 min–1
= 8.0 x 10–2 M min–1

(iii) In experiment IV

Rate = k[A]

Thus the completed table is

 exp. [A]/ mol L–1 [B]/M Initial rate of formation of D/mol L–1 min–1 I 0.1 0.1 2.0 x 10–2 II 0.2 0.2 4.0 x 10–2 III 0.4 0.4 8.0 x 10-2 IV 0.1 0.2 2.0 x 10–2

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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 âˆ†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

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

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 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