Rate of reaction = -1/2 d[NO₂]/dt
= K₁ [NO₂]² - K₂ [N₂O₄]
therefore rate of disapperance of NO₂
= - d[NO₂]/dt = 2K₁ [NO₂]² –2 K₂ [N₂O₄]

log K₂ /K₁ = ∆H/ 2.303R (1/T₁ –1/T₂)
log 6 /2 = ∆H/ 2.303R (1.5 x 10^-3 –2.0 x10^-3)
∆H comes negative. Hence exothermic.

Rate of reaction increases with temperature. Temperature coefficient is the ratio of rate constant at temperature (T + 10) K to the rate constant at temperature TK.
Temperature coefficient
\(=\frac { Rate \ constant \ at \ (T/10)K }{ Rate \ constanr \ at \ TK } \)
It is observed that for a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Unit of rate constant for first order,
Rate = \(\frac { dx }{ dt } =k[A]\)
\(\Rightarrow \frac { { mol \ L }^{ -1 } }{ s } =k \ (mol \ L^{ -1 })\)
K = s^-1

Factors influencing the rate of a chemical reaction are
Nature of reactants Different reactants require different amount of energies for breaking the old bonds and for the formation of new bonds. Hence, the reactivity of a substance is related to the ease with which the specific bonds are broken or formed
e.g. 2NO + O₂ \(\longrightarrow\) 2NO₂  ( fast )
2CO + O₂ \(\longrightarrow\)2CO₂   ( slow )
Concentration of reactants Rate of reaction is directly proportional to the concentration of the reactants.
Temperature Rate of reaction increases with increase in temperature.
Catalyst It alters the rate of reaction without being consumed in the reaction. It provides an alternative path to the reaction with a low energy barrier.

From the slow step, Rate = k₁ [NO₃] [NO]....(i)
From fast step, eqn. const. k = \(\frac { \left[ { NO }_{ 3 } \right] }{ \left[ NO \right] \left[ { NO }_{ 2 } \right] } \)....(ii)
Substituting the value of [NO₃] from (ii) in (i), we get: Rate = k' [NO]² [O₂]

Complex reaction proceeds through several elementary reactions.  Molecularity of each elementary reaction may be different, therefore, molecularity of complex reaction can't be determined.  Order of complex reaction is determined by slowest step in mechanism (involving elementary reactions).

t_1/2 = \(\frac{0.693}{k}\)\(\Rightarrow\)k₁ = \(\frac{0.693}{20}\)at 300 K,
k₂ = \(\frac{0.693}{5}\) at 350 K
log \(\frac { { K }_{ 2 } }{ { K }_{ 1 } } =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right) \)
\(log\left( \frac { 0.693 }{ 5 } \times \frac { 20 }{ 0.693 } \right) =\frac { { E }_{ a } }{ 2.303R } \left( \frac { 1 }{ { T }_{ 1 } } -\frac { 1 }{ { T }_{ 2 } } \right) \)
E_a = 2.303 x 8.314 x \(\frac { { T }_{ 1 }{ T }_{ 2 } }{ { T }_{ 2 }-{ T }_{ 1 } } \)log 4
= \(\frac { 19.147\times 350\times 300 }{ 50 } \times 0.6021\)
= 24.21 kJ mol^-1

It is because these molecules do not collide in a proper orientation that is why the reaction is slow.