Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Chemical Kinetics

List the factors on which the rate of a chemical reaction depends.

The various factors which affect the rate of chemical reactions are:
(i) Concentration of reactants
(ii) Temperature of reaction
(iii) Presence of catalyst
(iv) Nature of reactants
(v) Surface area
(vi) Exposure to radiations.

238 Views

For the reaction
: 2A + B + C → A₂B
the rate = k[A] [B]² with k = 2.0 x 10^–6 mol^–2L²s^–1.
Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1and [C] = 0.8 M. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^–1.

In the reaction
2A + B + C → A₂B + C,
there is no change in C, therefore its conc. does not affect the rate of the reaction.
Initial rate = k[A] [B]²But [A] = 0.1 M,
[B] = 0.2 M
and k = 2 x 10^–6 M^–2 s^–1
Therefore initial rate

Rate= [k] x [A] x [B]²

= 2 x 10^–6 M^–2 s^–1 x (0.1 M) (0.2 mol M)² = 8 x 10^–9 ms^–1From the equation:

2A + B + C → A₂B + C,

it is clear that when 2 moles of A are used then 1 mol of B is used in the same time. Therefore, when A has been reduced to 0.06 M (due to its 0.04 M has been reacted to 0.02 of B). Thus,

Conc. of A left = [A] = 0.06 M
Conc. of B left = [B] = [0.02 M – 0.02 M]
= 0.018 M

Rate = k[A] [B]²= 2 x 10^–6 M² S^–1 x (0.06 M) (0.18 M)
= 3.89 x 10^–9 Ms^–1.

263 Views

The decomposition of NH₃ on platinum surface is a zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 x 10^–4 mol^–1 L s^–1 ?

Rate of reaction can be

\frac{dx}{dt} = - \frac{1}{2} \frac{d ({NH}_{3})}{dt} = \frac{d (N_{2})}{dt} = \frac{1}{3} \frac{d [H_{2}]}{dt} = k .

k is rate constant and reaction is of zero order. Therefore, rate of reaction

\frac{dx}{dt} = \frac{d [H_{2}]}{dt} = 2.5 \times 10^{- 4} {Ms}^{- 1}

Rate of production of

H_{2},

is given by

\frac{d [H_{2}]}{dt} = \frac{3 d [H_{2}]}{dt} = 3 \times 2.5 \times 10^{- 4} {Ms}^{- 1} = 7.5 \times 10^{- 4} {Ms}^{- 1} .

883 Views

The decomposition of NH₃ on platinum surface is zero order. What are the rates of production of N₂ and H₂ if k = 2.5 x 10^–4 mol^–1Ls^–1?

The decomposition of NH₃ on plantinum surface is given as,

2 {NH}_{3} \to N_{2} + 3 H_{2} Rate of reaction, \frac{dx}{dt} = - \frac{1}{2} \frac{d ({NH}_{3})}{dt} = \frac{d [N_{2}]}{dt} = \frac{1}{3} \frac{d [H_{2}]}{dt} = k .

k is rate constant and reaction is of zero order.
Therefore, rate of reaction,

\frac{dx}{dt} = \frac{d [H_{2}]}{dt} = 2.5 \times 10^{- 4} {Ms}^{- 1}

Rate of production of

H_{2},

is given by

\frac{d [H_{2}]}{dt} = \frac{3 d [H_{2}]}{dt} = 3 \times 2.5 \times 10^{- 4} {Ms}^{- 1} = 7.5 \times 10^{- 4} {Ms}^{- 1} .

173 Views

From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants
(i) 3NO(g) → N₂O (g) Rate = k[NO]²
(ii) H₂O₂ (aq) + 3I^– (aq) + 2H⁺ → 2H₂O (l) + I^-₃Rate = k[H₂O₂][I^-]
(iii) CH₃CHO (g) → CH₄ (g) + CO(g) Rate = k [CH₃CHO]3/2
(iv) C₂H₅Cl (g) → C₂H₄ (g) + HCl (g) Rate = k [C₂H₅Cl]

(a) Rate = k[NO]²Order w.r.t. NO(g) = 2
Overall order of reaction = 2
Rate constant,

k = \frac{Rate}{[NO]^{2}} = \frac{conc . / time}{(conc .)^{2}}

\frac{1}{conc . time} = \frac{1}{mol L^{- 1} s} = L {mol}^{- 1} s^{- 1}

The dimensions of the rate constant, k are L mol^–1 s^–1(b) Rate = k[H₂O₂] [I^–]
Order w.r.t., H₂O₂ = 1
Order w.r.t., I^–= 1
Order w.r.t., I⁺ = 0
Overall order of reaction = 1 + 1 = 2
Rate constant,

k = \frac{Rate}{[H_{2} O_{2}] [I^{-}]} = \frac{conc . / time}{conc .^{2}}

= \frac{1}{conc . time} = \frac{1}{mol L^{- 1} s} = L {mol}^{- 1} s^{- 1}

The dimensions of rate constant, k are L mol^–1 s^–1.

(c) Rate = k[CH₃CHO]^3/2Order w.r.t., CH₃CHO = 3/2 = 1.5
Overall order of reaction = 1.5
Rate constant,
$k = \frac{Rate}{[{CH}_{3} CHO]^{3 / 2}} = \frac{conc . / time}{conc .^{3 / 2}} = \frac{1}{conc .^{1 / 2} \times s} = \frac{1}{(mol L^{- 1})^{1 / 2} s} = L^{1 / 2} {mol}^{- 1 / 2} s^{- 1}$
The dimensions of rate constant, k are L^1/2mol^–1/2 s^–1.

(d) Rate = $k [{CHCl}_{3}] [{Cl}_{2}]^{1 / 2}$
Order w.r.t. $[{CHCl}_{3}] = 1$
Order w.r.t. ${Cl}_{2} (g) = \frac{1}{2} = 0.5$
Overall order of reaction = 1+0.5 = 1.5
Rate constant,
$k = \frac{Conc . / time}{[{CHCl}_{3}] {[{Cl}_{2}]}^{1 / 2}} = \frac{conc . / time}{conc .^{3 / 2}} = \frac{1}{(mol L^{- 1})^{1 / 2} x s} = L^{1 / 2} {mol}^{- 1 / 2} s^{- 1}$
The dimensions of rate constant, k are L^1/2mol^–1/2 s^–1.

205 Views

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

Chemical Kinetics