﻿ What do you mean by zero order reaction? How the value of rate constant is determined? What is the relation between rate constant and half-life period? from Chemistry Chemical Kinetics Class 12 Nagaland Board

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What do you mean by zero order reaction? How the value of rate constant is determined? What is the relation between rate constant and half-life period?

When the rate of the reaction is independent of the concentration of the reactants, the reaction is known as zero order reaction. In zero order reaction, the concentration of reactant (R) remains unaltered during the course of reaction.

It means
or

on integration of this equation, we get

...(i)

Since

where t = 0 (i.e., no product is formed at the beginning of the reaction), the constant must be zero.

Thus, [R]° = – K x 0 = constant
Constant = [R]°
Put the value of constant in eqn. (i), we get

[R] = – kt + [ R]°                                 ...(ii)

When a plot is drawn between concentration of reactant (R) and time t, a straight line is obtained, slope of which gives the value of – k and intercept on y-axis is equal to the value of [R]°.
Alternatively, the value of k can be obtained from eqn. (i) by putting the value of concentration, [R] at any time t and initial concentration, [R]°.
Hence,   $\mathrm{k}=\frac{\left[\mathrm{R}\right]°-\left[\mathrm{R}\right]\mathrm{t}}{\mathrm{t}}$
Half life period is the time period to reduce the initial concentration of the reactant to half of its initial value.
Thus,
Thus, the half life period is directly proportional to the initial concentration of the reactant, [R]° and inversely proportional to the rate constant k.
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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 ∆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

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

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

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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