﻿ Methyl acetate is hydrolysed in approximately N-HCl at 25°C. 5.0 mL portions of the reaction mixture were removed at intervals and titrated with 0.185 N-NaOH. From the data given below prove that hydrolysis of methyl acetate is a first order reaction. t/sec 1242 sec 2745 sec 4546 sec At Conc. -27.80ml -29.70ml -31.81ml from Chemistry Chemical Kinetics Class 12 Nagaland Board

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Methyl acetate is hydrolysed in approximately N-HCl at 25°C. 5.0 mL portions of the reaction mixture were removed at intervals and titrated with 0.185 N-NaOH. From the data given below prove that hydrolysis of methyl acetate is a first order reaction.
 t/sec 1242 sec 2745 sec 4546 sec At Conc. -27.80ml -29.70ml -31.81ml

Here [ A]0 is proportional to the alkali consumed between t = &&& and t = 339 sec, i.e., [A]0 = (39.81 mL – 26.34 mL) = 13.47 mL [A]t is proportional to the alkali consumed at &&& minus the alkali consmed at the specified time. Thus,
[A], at 1242 sec = 39.81 mL – 27.80 mL = 12.01 mL
[A], at 2745 sec = 39.81 mL – 29.70 mL = 10.11 mL
[A], at 4546 sec = 39.81 mL – 31.81 mL = 8.00 mL
Substituting the values of t, [A]0 and [A]t in the first-order rate equation

we get the following values of k

(i)

(ii)

(iii)

Since the three values of k derived from first-order reaction are close to being identical, hydrolysis of methyl acetate is a first order reaction.

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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

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

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