$$R \to P$$
For a second order reaction, rate $$ = k{\left[ R \right]^2}$$
If conc. of $$R$$  is increased by four times, rate $$ = k{\left[ {4R} \right]^2}$$
Hence, the rate of formation of $$P$$  increases by 16 times.

Units of $$k$$ indicate that reaction (i) is of second order and reaction (ii) is first order.
For (i) reaction, $${t_{\frac{1}{2}}} \propto \frac{1}{a},$$
$$\eqalign{ & {\rm{first}}\,\,{t_{{1 \over 2}}} = 1\,hr,\,{\rm{second}}\,{t_{{1 \over 2}}} = 2\,hr \cr & \left[ A \right] = 1M\buildrel {1\,hr} \over \longrightarrow 0.5M\buildrel {2\,hr} \over \longrightarrow 0.25M \cr & \left[ B \right] = 1M\buildrel {1\,hr} \over \longrightarrow 0.5M\buildrel {1\,hr} \over \longrightarrow 0.25M\buildrel {1\,hr} \over \longrightarrow 0.12M \cr & {{\left[ A \right]} \over {\left[ B \right]}} = {{0.25M} \over {0.125M}} = 2 \cr} $$

\[\begin{align} & {{\left[ A \right]}_{0}}=2.0\,m,\,\left[ A \right]=0.15\,m,\,t=200\,\min \\ & \text{For first order reaction} \\ & {{A}_{0}}=\text{Initial concentration} \\ & A=\text{Final concentration} \\ & \text{Rate constant,} \\ & k=\frac{2.303}{t}\text{log}\frac{{{\left[ A \right]}_{0}}}{\left[ A \right]} \\ & \,\,\,\,\,=\frac{2.303}{200}\text{log}\frac{2.0}{0.15} \\ & \,\,\,\,\,=\frac{2.303}{200}\left( \text{log}\,200-\text{log}\,15 \right) \\ & \,\,\,\,\,=\frac{2.303}{200}\times \left( 2.3010-1.1761 \right) \\ & \,\,\,\,\,=\frac{2.303\times 1.1249}{200} \\ & \,\,\,\,\,=0.01295\,{{\min }^{-1}} \\ & \text{Now, half life,} \\ & {{t}_{\frac{1}{2}}}=\frac{0.6932}{k} \\ & \,\,\,\,\,\,\,=\frac{0.6932}{0.01295} \\ & \,\,\,\,\,\,\,=53.50\,\min \\ \end{align}\]

A catalyst does not take part in chemical reaction but it alters the rate of reaction.

The activation energy of reverse reaction will depend upon whether the forward reaction is exothermic or endothermic.
As $$\Delta H = {E_a}$$   ( forward reaction ) $$ - {E_a}$$  ( backward reaction )
$$\eqalign{ & {\text{For exothermic reaction}} \cr & \Delta H = - ve \cr & \therefore \,\, - \Delta H = {E_a}\left( f \right) - {E_a}\left( b \right) \cr & {\text{or}}\,\,{E_a}\left( f \right) = {E_a}\left( b \right) - \Delta H \cr & \therefore \,\,{E_a}\left( f \right) < {E_a}\left( b \right) \cr & {\text{for endothermic reaction}} \cr & \Delta H = + ve \cr} $$
$$\therefore \,\,\Delta H = {E_a}\left( f \right) - {E_a}\left( b \right)$$      $${\text{or}}\,\,{E_a}\left( f \right) = \Delta H + {E_a}\left( b \right)$$
$$\therefore \,\,{E_a}\left( f \right) > {E_a}\left( b \right).$$

The given reaction is $$:2X + Y \to Z$$
$$ - \,\frac{{d\left[ X \right]}}{{2dt}} = \frac{{d\left[ Z \right]}}{{dt}}$$
∴ Rate of formation of $$Z$$ is half of the rate of disappearance of $$X.$$

Key Idea In endothermic reactions, energy of reactants is less than that of the products.
Potential energy diagram for endothermic reactions is

where,
$${E_a} = $$  activation energy of forward reaction
$${{E'}_a} = $$  activation energy of backward reaction
$$\Delta H = $$   enthalpy of the reaction.
From the above diagram,
$$\eqalign{ & {E_a} = {{E'}_a} + \Delta H \cr & {\text{Thus,}}\,\,{E_a} > \Delta H \cr} $$

Energy of activation is lowered when a catalyst is used.

Activation energy can be calculated by using Arrhenius equation. The Arrhenius equation is
$$\log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303R}}\left[ {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$$
where, $${k_1}$$  and $${k_2} = $$  rate constants at two different temperatures, i.e. $$\,{T_1}$$  and $${T_2}$$  respectively.
$${E_a} = $$  Activation energy
$$R=$$  Gas constant
So, activation energy of a chemical reaction can be determined by evaluating rate constants at two different temperatures.

Activation energy of reactant is less than the energy of activation of products.