﻿ The gas phase decomposition of acetaldehydeCH3CHO(g) → CH4(g) + CO(g)at 680 K is observed to follow the rate expressionrate=-dCH3CHOdt=kCH3CHO1/2If the rate of the decomposition is followed by monitoring the partial pressure of the acetaldehyde, we can express the rate asdpCH3CHOdt=kpCH3CHO3/2If the pressure is measured in atmosphere and time in minutes then(i) What are the units of the rate of reaction?(ii) What are the units of rate constant k? from Chemistry Chemical Kinetics Class 12 Nagaland Board

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The gas phase decomposition of acetaldehyde
CH3CHO(g) → CH4(g) + CO(g)
at 680 K is observed to follow the rate expression
$\mathrm{rate}=\frac{-\mathrm{d}\left[{\mathrm{CH}}_{3}\mathrm{CHO}\right]}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{CH}}_{3}\mathrm{CHO}\right]}^{1/2}$
If the rate of the decomposition is followed by monitoring the partial pressure of the acetaldehyde, we can express the rate as
$\frac{{\mathrm{dp}}_{{\mathrm{CH}}_{3}\mathrm{CHO}}}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{p}}_{{\mathrm{CH}}_{3}\mathrm{CHO}}\right]}^{3/2}$
If the pressure is measured in atmosphere and time in minutes then
(i) What are the units of the rate of reaction?
(ii) What are the units of rate constant k?

The gas phase decomposition of acetaldehyde

CH3CHO(g) → CH4(g) + CO(g)

at 680 K is observed to follow the rate expression
$\mathrm{rate}=\frac{-\mathrm{d}\left[{\mathrm{CH}}_{3}\mathrm{CHO}\right]}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{CH}}_{3}\mathrm{CHO}\right]}^{1/2}$

Thus,

(i) Units of rate of reaction =

(ii)

108 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

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

2088 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