﻿ The rate law for the gas phase reaction of chlorform with chlorine.CHCl3(g) + Cl2(g) → CCl 4(g) + HCl(g)is given by rate = k[CHCl3] [Cl2]1/2. How would the rate of reaction vary when (a) the concentration of CHCl3 is doubled (b) the concentration of Cl2 is doubled. What is the effect of each of these two changes on the rate constant. from Chemistry Chemical Kinetics Class 12 Nagaland Board

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The rate law for the gas phase reaction of chlorform with chlorine.
CHCl3(g) + Cl2(g) → CCl 4(g) + HCl(g)
is given by rate = k[CHCl3] [Cl2]1/2. How would the rate of reaction vary when (a) the concentration of CHCl3 is doubled (b) the concentration of Cl2 is doubled. What is the effect of each of these two changes on the rate constant.

The rate law for the gas phase reaction of chlorform with chlorine.
CHCl3(g) + Cl2(g) → CCl 4(g) + HCl(g)
the rate is given by

rate1 =k[CHCl3] [Cl2]1/2 .....1

if the concentration of CHCl3 is doubled
then

rate2 =k[CHCl3]2 [Cl2]1/2 .....2
divide 1by 2 we get

if the concentration of Cl2  is doubled
then .
rate3 =k[CHCl3] [Cl2]2/2........3

divide 1 by 3
we get

thus the rate reaction is double in both case.

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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

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.

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

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