It is know that,
\(t=\frac { 2.303 }{ k } \log { \frac { { \left[ R \right] }_{ 0 } }{ \left[ R \right] } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log\frac { 1 }{ \frac { 1 }{ 16 } } \)
= \(\frac { 2.303 }{ 60{ s }^{ -1 } } \log16\)
= 4.6 x 10^-2 s (approximately)
Hence the required time is 4.6 x 10^-2 s

(i) The differential rate equation will be 
\(\frac { -d\left[ R \right] }{ dt } \) = k[A][B]²
(ii) if the concentration of B is increased three times, then
\(\frac { -d\left[ R \right] }{ dt } \) = k[A] [3B]²
= 9.k [A] [3B]²
∴ The rate of reaction will increase 9 times.
(iii) When the concentrations of both A and B are doubled
\(\frac { -d\left[ R \right] }{ dt } \) =k [A] [B]²
= k [2A] [2B]²
= 8.k [A] [B]²
∴ The rate of reaction will increase 8 times.

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 \({ \frac { 1 }{ 4 } }^{ th }\)

(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

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