(i) The rate constant is nearly doubled with a rise in temperature by 10^oC for a chemical reaction.
(ii) The temperature effect on the rate constant can be represented quantitatively by Arrhenius equation.
\(k=A{ e }^{ \frac { -{ E }_{ a } }{ RT } }\)

Let the concentration of the reactant be [A] = a
Rate of reaction R = k [A]²
R=ka²
(i) R = k (2a)²
= 4 ka²
= 4R
∴ The rate of the reaction would increase by 4 times.
(ii) If the concentration of the reactant is reduced to half [A] = \(\frac { 1 }{ 2 } \) a than the rate of reaction would be
\(R=k\left( \frac { 1 }{ 2 } a \right) 2\)
\(=\frac { 1 }{ 4 } k{ a }^{ 2 }\)
\(=\frac { 1 }{ 4 } kR\)
∴ The rate of the reaction would reduced by \({ \frac { 1 }{ 4 } }^{ th }\)

\(2{ NH }_{ { 3 }_{ (g) } }\overset { pt }{ \longrightarrow } { N }_{ { 2 }_{ (g) } }+3{ H }_{ { 2 }_{ (g) } }\)
Rate = \(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } \)
However, it is given that the reaction is of zero order. Therefore
\(-\frac { 1 }{ 2 } \frac { d\left[ { NH }_{ 3 } \right] }{ dt } =\frac { d\left[ { N }_{ 2 } \right] }{ dt } =\frac { 1 }{ 3 } \frac { d\left[ { N }_{ 2 } \right] }{ dt } k\)
= 2.5 x 10^-4 mol L^-1 S^-1
The rate of production of N₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)=  2.5 x 10^-4 mol L^-1 S^-1
The rate of production of H₂ is
\(\frac { d\left[ { N }_{ 2 } \right] }{ dt } \)= 3 x 2.5 x 10^-4 mol L^-1 s^-1
= 7.5 x 10^-4 mol L^-1 s^-1.

(i) Given rate = K[NO]² 
∴ Order of the reaction = 2
\(K=\frac { Rate }{ { \left[ NO \right] }^{ 2 } } \)
Dimension of = \(\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ { \left( mol{ L }^{ -1 } \right) }^{ 2 } } \)
\(=\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ { mol }^{ 2 }{ L }^{ -2 } } \)
= L mol^-1 S^-1
(ii) Given rate = K[H₂O₂][I^-]
∴ order of reaction = 2
\(K=\frac { Rate }{ \left[ { H }_{ 2 }{ O }_{ 2 } \right] \left[ { I }^{ - } \right] } \)
Dimension of = \(\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ \left( mol{ L }^{ -1 } \right) \left( mol{ L }^{ -1 } \right) } \)
= L mol^-1 S^-1
(iii) Given rate = K[CH₃CHO]^3/2
∴ Order of the reaction = 3/2
\(K=\frac { Rate }{ { \left[ { CH }_{ 3 }CHO \right] }^{ \frac { 3 }{ 2 } } } \)
Dimension of = \(\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ { \left( mol{ L }^{ -1 } \right) }^{ \frac { 3 }{ 2 } } } \)
\(=\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ { mol }^{ \frac { 3 }{ 2 } }{ L }^{ -\frac { 3 }{ 2 } } } \)
\(={ L }^{ \frac { 1 }{ 2 } }{ mol }^{ -\frac { 1 }{ 2 } }{ S }^{ -1 }\)
Given rate = K[C₂H₅Cl]
∴ order of the reaction = 1
\(K=\frac { Rate }{ \left[ { C }_{ 2 }{ H }_{ 5 }CL \right] } \)
Dimension of \(=\frac { mol{ L }^{ -1 }{ S }^{ -1 } }{ { mol }{ L }^{ -1 } } \) = -1

In the given case
E_a = 209.5 KJ mol^-1 = 209500 J mol^-1
T = 581 K
R = 8.314 JK^-1 mol^-1
Now the fraction of molecules of reactants having energy equal to or greater than activation energy is given as.
x = e^-E_a/RT
\(\Rightarrow \log { x } =-\frac { { E }_{ a } }{ RT } \)
\(\Rightarrow \log { x } =-\frac { { E }_{ a } }{ 2.303RT } \)
\(\Rightarrow \log { x } =\frac { 209500J{ mol }^{ -1 } }{ 2.303\times 8.314J{ K }^{ -1 }{ mol }^{ -1 }\times 581 } \)
= 18.8323
x = Antilogs (18.8323)
= Antilogs \(\overset { \_ \_ }{ 19 } .1677\)
= 1.471 x 10^-19

Given:
K₁ = 1.60 x 10^-5 S^-1; T₁ = 600K
K₂ = 6.36 x 10^-3 S^-1; T₂ = 700K
We know that log \(\frac { { k }_{ 2 } }{ { k }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left[ \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right] \)
Substituting the values of K₁, K_2, and R= (8.314 JK^-1 mol^-1), we get,
\(log\frac { 6.36\times { 10 }^{ -3 } }{ 1.60\times { 0 }^{ -5 } } =\frac { { E }_{ a } }{ 2.303\times 8.314 } \left[ \frac { 1 }{ 600 } -\frac { 1 }{ 700 } \right] \)
∴ E_a = 2.6 x 19.15 x 4200 J mol^-1
= 2.09 x 10⁵ J mol^-1
= 209 kJ mol^-1

For first order reaction
\(k=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
For t = t_1/2, \(x=\frac { a }{ 2 } \)
\({ t }_{ \frac { 1 }{ 2 } }=\frac { 2.303 }{ k } \log { \frac { a }{ a-\frac { a }{ 2 } } } \)
\(\frac { 2.303 }{ k } \log2=\frac { 2.303\times 0.3010 }{ 7.39\times { 10 }^{ -5 } } \)
= 9.38 x 10³ s.

Formula:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { E }_{ a }\left( { T }_{ 2 }-{ T }_{ 1 } \right) }{ 2.303R{ T }_{ 1 }{ T }_{ 2 } } \)
Given:
Energy of activation: E = 150 kJ = 150000 J
Temperatures: T₁ = 673 K; T₂ = 773 K
Rate constant: k₁ = 2.3 x 10^-3
Gas constant: R = 8.314 J k^-1 mol^-1
Solution:
\(\log { \frac { { k }_{ 2 } }{ { k }_{ 1 } } } =\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } =\frac { 150000(773-673) }{ 2.303\times 8.314\times 673\times 773 } \)
\(=\frac { 15000000 }{ 2.303\times 8.314\times 673\times 773 } \)
\(\log { \frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } } \) = 1.505
\(\frac { { k }_{ 2 } }{ 2.3\times { 10 }^{ -3 } } \) Antilog 1.5905 = 32
∴ k₂ = 2.3 x 10^-3 x 32 = 7.36 x 10^-2
k₂ = 7.36 x 10^-2

Formula:
k₁ = A.e^-E_a/RT
Given:
Rate constant, k₁ - 3.0 x 10^-4 s^-1
Frequency factor, A = 6.0 x 10¹⁴ s^-1
Activation energy, E_a = 104.4 kJ
= 104400 J
Temperatures, T₁ = 25^o C; T₂ =∞
Solution:
\({ k }_{ 2 }=A.{ e }^{ -\frac { { E }_{ a } }{ RT } }\)
k₂ = 6.0 x 10¹⁴ x \(\frac { 104400 }{ { e }^{ 8.314\times \infty } } \)
k₂ = 6.0 x 10¹⁴ x 1   \(\left( \therefore e\frac { -{ { E }_{ a } } }{ R\infty } =1 \right) \)
∴ The rate constant at T ⟶ ∞ is 6.0 x 10¹⁴ s^-1

(i) When the concentration of the reactant is increased by 'n' times, the rate of reaction is also increased by n times. That is, if the concentration of the reactant is doubled, the rate is doubled.
(ii) The unit of rate constant of a first order reaction is sec^-1 or time^-1.
\({ k }_{ 1 }=\frac { rate }{ (a-x) } =\frac { { mol.lit }^{ -1 }{ sec }^{ -1 } }{ { mol.lit }^{ -1 } } \)
(iii) The time required to complete a definite fraction of reaction is independent of the initial concentration, of the reactant if t_1/u is the time of one 'u' th fraction of reaction to take place then from equation.
\({ k }_{ 1 }=\frac { 2.303 }{ t } \log { \frac { a }{ a-x } } \)
\(x=\frac { a }{ u } and\quad { t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { a }{ a-\frac { a }{ u } } } ;\)
\({ t }_{ \frac { 1 }{ u } }=\frac { 2.303 }{ { k }_{ 1 } } \log { \frac { u }{ (u-1) } } \)
since k₁ = rate constant, t_1/u is independent of initial concentration 'a'.

(i) All radioactive transformations follow first order kinetics. For example,
₉₂U²³⁸ ⟶ ₉₀U²³⁴ +₂He⁴
(ii) Decomposition of sulphuryl chloride in the gas phase proceeds by first order kinetics.
SO₂Cl_2(g) ⟶ SO_2(g) + Cl_2(g)
(iii) Inversion of sucrose in acidic aqueous medium follows first order reaction.
C₁₂H₂₂O₁₁ +H₂O \(\overset { H+ }{ \longrightarrow } \) C₆H₁₂O₆ +C₆H₁₂O₆

A reaction is second order in A and first order in B
(i) Differential rate equation,
Rate = \(\frac { -d\left[ R \right] }{ dt } =k{ \left[ A \right] }^{ 2 }\left[ B \right] \)
(ii) When the concentration of A is increased three times, (i.e) 3A, then
Rate = k[3A]² [B]
= 9k [A]²[B] = 9 (initial rate)
This shows the rate will increase 9 times to the initial time.
(iii) When concentration of both A and B is doubled then,
Rate = k [2A]² [2B] = 8k [A]² [B] = 8 (initial rate)
This shows that rate will increase 8 times to the initial rate.

(i) (a) Molecularity of the reaction is the number of molecules taking part in an elementary step.
(b) For this we require at least a single molecule leading to the value of minimum molecularity of one.
(c) Hence, molecularity of any reaction can never be equal to zero.
(ii) For elementary reaction, (i.e) the reaction which proceeds in a single step, order, and molecularity have the same value.

\({ t }_{ \frac { 3 }{ 4 } }=\frac { 2.303 }{ K } \log { \frac { { \left[ R \right] }_{ 0 } }{ \frac { 1 }{ 4 } { \left[ R \right] }_{ 6 } } } \)
\({ t }_{ \frac { 3 }{ 4 } }=\frac { 2.303 }{ K } \log 4\)
\(=\frac { 2.303\times 0.6021 }{ K } =\frac { 1.386 }{ K } \)
\(=2\times \frac { 0.693 }{ K } \)
\(=2{ t }_{ \frac { 1 }{ 2 } }\)

\(k=\frac { 2.303 }{ t } \log\frac { \left( { V }_{ \infty }-{ V }_{ 0 } \right) }{ \left( { V }_{ \infty }-{ V }_{ 0 } \right) } \)
\(k=\frac { 2.303 }{ 20 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.1151 \log\frac { 30.2 }{ 24.8 } \)
= 0.1151 log 1.2479
= 0.1151 x 0.0959
= 11.03 x10^-3 min^-1
When t = 40 mts
\(k=\frac { 2.303 }{ 40 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.0576\times \log\frac { 30.2 }{ 20.9 } \)
= 9.19 x 10^-3 min^-1
When t = 60 mts
\(k=\frac { 2.303 }{ 60 } \log\frac { 50.4-20.2 }{ 50.4-25.6 } \)
\(=0.03838\times \log\frac { 30.2 }{ 17.6 } \)
= 0.03838 x 0.2343
= 8.99 x 10^-3 min^-1
The constant values of K show that the reaction is of first order.