ï»¿ The reaction:CH3COF + H2O → CH3COOH+HFhas been studied under the following initial conditions.CASE I:                                                 CASE II:CH2O = 1.00 M                                        CH2O = 0.02 M.CCH3COF = 0.01 M                                 CCH3COF = 0.80 M                    Concentrations were monitored as a function of time and are given below:Determine the order of the reaction and the rate constant for the reaction. from Chemistry Chemical Kinetics Class 12 Nagaland Board

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The reaction:

has been studied under the following initial conditions.

Concentrations were monitored as a function of time and are given below:

Determine the order of the reaction and the rate constant for the reaction.

Let rate k(CCH3COF)(CH2O)b
CH2O >> CCH3COF and in the second case, CH2O << CCH3COF. In the first case we determine the order of the reaction with respect to CH3 COF. We note that the reaction is not of zero order as rate of reaction changes with time.

Therefore k(CH2O) = 0.0154 min–1 and we note the order of reaction with respect to CH3COF is 1.
Now we determine the order of raction with respect to water.
Again,

The reaction is first order in H
2O and we have

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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

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

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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

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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.

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