(i) The slowest step among the elementary  reactions determines the rate of reaction, e.g.
\({ NO }_{ 2 }+{ F }_{ 2 }\overset { Slow }{ \longrightarrow } { NO }_{ 2 }F+F\)
\({ NO }_{ 2 }+F\overset { Fast }{ \longrightarrow } { NO }_{ 2 }F\)
From slow step
\(Rate=k\left[ { NO }_{ 2 } \right] \left[ { F }_{ 2 } \right] \)
(ii) Molecularity is defined as number of atoms or molecules or ions which must collide to each other simultaneously as to result in chemical reaction, e.g.
\({ H }_{ 2 }(g)+I_{ 2 }(g)\longrightarrow 2HI(g)\)
It is a biomolecular reaction.

\(k=\frac { 2.303 }{ t } \log { \frac { \left[ { R }_{ 0 } \right] }{ \left[ R \right] } }\)
\( \\ \Rightarrow \ 2.4\times { 10 }^{ -3 }{ s }^{ -1 }=\frac { 2.303 }{ { t }_{ { 3 }/{ 4 } } } \log { \frac { \left[ { R }_{ 0 } \right] }{ \frac { 1 }{ 4 } \left[ { R }_{ 0 } \right] } }\)
\({ t }_{ { 3 }/{ 4 } }=\frac { 2.303 }{ 2.4\times { 10 }^{ -3 } } \log { 4 } =\frac { 2.303\times 0.6021 }{ 2.4\times { 10 }^{ -3 } } \)
\({ t }_{ { 3 }/{ 4 } }=\frac { 1.386\times { 10 }^{ 3 } }{ 2.4 } =\frac { 13.86{ \times 10 }^{ 2 } }{ 2.4 }\)
\( \\ { t }_{ { 3 }/{ 4 } }=5.77{ \times 10 }^{ 2 }s\)

It is because multimolecular collisions have least or rare probability.

It is because activation energy is higher and therefore reaction is not feasible at room temperature.

Energy of activation is the extra energy which must be supplied to the reactants so that these can change into products. It is difference between energy of reactants and energy of activated complex as shown in diagram below. 

The reactants molecules which possess activation energy and collide in proper orientation lead to formation of product molecules.

Activated complex is intermediate compound between reactants and products as shown above. It is highly unstable as it has highest energy. It readily charges into product. Those molecules which can form activated complex can lead to formation of products, e.g

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

Rate of reaction increases with increase in temperature because kinetic energy of molecules increases, increases, number of molecules possessing activation energy increases, rate of reaction increases.

\(\frac{a}{t}=\mathrm{k} \Rightarrow \mathrm{t}=\frac{a}{k}\)  In time \(\frac{a}{k}\) the reaction will go to completion in case of zero order reaction.

Rate constant is not affected by reactants' concentrations.

(d) : For zero order reaction, k = \(\frac {1}{t}\)\(\left[ { \left[ A \right] }_{ 0 }-\left[ A \right] \right] \) or \( { \left[ A \right] }-\left[ A \right] _{ 0 } - kt\)

(a) : The rate of constant of a reaction depends only on temperature and does not depend upon concentrations of the reactants.

(d) \(\frac{1}{3} \frac { d[{ Br }_{ 2 }] }{ dt } =-\frac { 1 }{ 5 } \frac { d[{ Br }^{ - }] }{ dt } \) or \(\frac { d[{ Br }_{ 2 }] }{ dt } =-\frac {3 }{ 5 } \frac { d[{ Br }^{ - }] }{ dt } \)

(i) As pressure of X is changing with time, it cannot be a zero order reaction. Let us now check it for 1st order.
At t = 100 min, \(k={2.303\over 100}log{P_0\over P_t}={2.303\over 100}log {800\over 400}=6.932\times 10^{-3}min^{-1}\)
At t = 200 min, \(k={2.303\over 200}log{800\over 200}={2.303\over 800}log4=6.932\times10^{-3}min^{-1}\)
As k comes out to be constant, hence it is a reaction of 1st order
(ii)\(t_{75./.}={2.303\over k}log{100\over 100-75}={2.303\over 6.932\times10-3min}log4=200min\)
(iii) 2x (g)⟶ 3 y (g) + 2z (g)
Initial Pressure             800mm      0             0
Pressure after time t     800 - 2p    3 p           2 p
When pressure of X is 700 mm, 800 - 2 p = 700 or p = 50 mm
Total pressure = (800 - 2 p) + 3 p + 2 p = 800 + 3 p = 800 + 3 x 50 = 950 mm.