Chemical Kinetics

Chemistry I

Chemistry

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The rate of change in concentration of C in the reaction 2A + B → 2C + 3D was reported 1.0 M sec^{–1}. Calculate the reaction rate as well as rate of change of concentration of A, B and D.

Rate of reaction =

$-\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=-\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}$

$\because $ $\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=1.0\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}$

$\therefore $

$-\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=1.0\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}\phantom{\rule{0ex}{0ex}}-\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=0.5\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{d}\left[\mathrm{D}\right]}{\mathrm{dt}}=\frac{3}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}=\frac{3}{2}\times 1=1.5\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}\phantom{\rule{0ex}{0ex}}$

Also, Because $\mathrm{Rate}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}$

$\therefore $ $\mathrm{Rate}=\frac{1}{2}\times 1=0.5\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{sec}}^{-1}.$

Dependence of rate on concentration of reactants: The rate of a chemical reaction at a given temperature may depend on the concentration of one or more reactants and sometimes on products. The representation of rate of a reaction in the terms of the concentration of the reactants is given by rate law. The rate for a given reaction is established by experimental study of the rate of reaction over a wide range of concentration of the reactants and products. Rate law expression differs for the same reaction under different experimental conditions. Rate constants and order of reaction:

(i) Rate constant or specific reaction rate: It is the rate of reaction when the concentration of each reactant is 1 mol/L. For a given reaction it is constant at a particular temperature and is independent of the concentration of reactants. The units of the rate constant of a reaction depends on the order of the reaction. For an nth order of reaction,

$\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{conc}{)}^{\mathrm{n}}\phantom{\rule{0ex}{0ex}}\mathrm{k}=\frac{\mathrm{dx}}{\mathrm{dt}}\times \frac{1}{(\mathrm{conc}{)}^{\mathrm{n}}}=\frac{\mathrm{conc}.}{\mathrm{time}}\times \frac{1}{(\mathrm{conc}{)}^{\mathrm{n}}}=\frac{1}{\mathrm{time}}\times \frac{1}{(\mathrm{conc}.{)}^{\mathrm{n}-1}}$

For zero order of reaction, units of k is mol L^{–1} time^{–1}, for first order reaction, unit of k is time^{–1}, for second order reaction, unit of k is L mol^{–1} time^{–1}.

In terms of gaseous reactions, concentration is expressed terms of pressure having units of atmosphere. Let us consider the general reaction: aA + bB → Products

where A and B are the reactants and a and b are the stoichiometric coefficients in the balanced chemical equations.

The rate law is written as,

Rate = Δ[A] / Δ t = k[A]^{α} [B]^{β}where k is called the rate constant. Rate constant (k) is the constant of proportionality within the empirical rate law linking the rate of reaction and concentration of reactants involved in the reaction. The rate law can be written in the form

Rate ∝ [A]^{α}[B]^{β}

The exponents ‘α’ and ‘β’ in the rate law indicate how sensitive the rate is to change in [A] and [B] and they are usually unrelated to the coefficients a and b in the balanced equation. In general, exponents are positive. But for complex reactions it can be negative, zero or even fractions. If exponent is one, it means rate depends linearly on the concentration of the reactant. If concentration of A is doubled, rate is also doubled. This means a = 1. If α = 2 and [A] is also doubled, rate increases by the factor of 4(z^{2}). When exponent is zero {[A]^{0} = 1}, rate is independent of concentration.

165 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 [A

Final concentration [A

Time is = 10 min

Rate of reaction = Rate of disappearance of A.

Rate of reaction = $-\frac{1}{2}\frac{\u2206\left[\mathrm{A}\right]}{\u2206\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

$=-\frac{1}{2}\frac{(0.4-0.5)\mathrm{mol}{\mathrm{L}}^{-1}}{10\mathrm{minute}}\phantom{\rule{0ex}{0ex}}=\frac{0.1\mathrm{mol}{\mathrm{L}}^{-1}}{5\mathrm{minutes}}\phantom{\rule{0ex}{0ex}}=0.005\mathrm{mol}{\mathrm{litre}}^{-1}{\mathrm{min}}^{-1}.$

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]

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]

Final quantity, [R] = 3 g

Rate constant, k = 1.15 x 10

Formula of 1

We know that

$\mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}$

or

$\mathrm{t}=\frac{2.303}{1.15\times {10}^{-3}}\mathrm{log}\left(\frac{5}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(\mathrm{log}5-\mathrm{log}3)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(0.6990-0.4771)\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219}{1.15\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219\times 1000}{1.15}\phantom{\rule{0ex}{0ex}}=444\mathrm{sec}.$

$\mathrm{t}=\frac{2.303}{1.15\times {10}^{-3}}\mathrm{log}\left(\frac{5}{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(\mathrm{log}5-\mathrm{log}3)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2.303}{1.15\times {10}^{-3}}(0.6990-0.4771)\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219}{1.15\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=\frac{2.303\times 0.2219\times 1000}{1.15}\phantom{\rule{0ex}{0ex}}=444\mathrm{sec}.$

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, R_{1} = k .(a)^{2}

= ka^{2}

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, R_{2} = k (3a)^{2}

= 9ka^{2}

We have already get that R_{1} = ka_{2} plus this value we get

R_{2} = 9 R_{1}

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 A^{2}Thus, rate will increase 9 times.

972 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 âˆ†t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )

The formula of average rate of change

${\mathrm{r}}_{\mathrm{av}}=\frac{-\u2206\mathrm{R}}{\u2206\mathrm{t}}=\frac{\u2206\left[\mathrm{P}\right]}{\u2206\mathrm{t}}$

(i) Average rate

$=\frac{(0.03-0.02)\mathrm{M}}{25\times 60\mathrm{sec}}\phantom{\rule{0ex}{0ex}}=\frac{0.01\mathrm{M}}{25\times 60\mathrm{s}}=6.66\mathrm{M}{\mathrm{s}}^{-1}$

(ii) Average rate

$=\frac{(0.03-0.02)\hspace{0.17em}\mathrm{M}}{25\mathrm{min}}=\frac{0.01\mathrm{M}}{25}\phantom{\rule{0ex}{0ex}}=0.0004{\mathrm{Ms}}^{-1}.$

(i) Average rate

$=\frac{(0.03-0.02)\mathrm{M}}{25\times 60\mathrm{sec}}\phantom{\rule{0ex}{0ex}}=\frac{0.01\mathrm{M}}{25\times 60\mathrm{s}}=6.66\mathrm{M}{\mathrm{s}}^{-1}$

(ii) Average rate

$=\frac{(0.03-0.02)\hspace{0.17em}\mathrm{M}}{25\mathrm{min}}=\frac{0.01\mathrm{M}}{25}\phantom{\rule{0ex}{0ex}}=0.0004{\mathrm{Ms}}^{-1}.$

1717 Views