A catalyst provides an alternate path to the reaction which has lower activation energy.

At low pressure, rate is proportional to the surface coverage and is of first order while at high pressure it follows zero order kinetic due to complete coverage of surface area.

For the first order reaction

A $$ \to $$ Product

Rate $$ \propto $$ [A]

k = $${{2.303} \over {{t_2} - {t_1}}}\log {{{{\left( {rate} \right)}_1}} \over {{{\left( {rate} \right)}_2}}}$$

= $${{2.303} \over {20 - 10}}\log {{0.04} \over {0.03}}$$

= 0.0287 sec^-1

$${t_{{1 \over 2}}} = {{0.693} \over k} = {{0.693} \over {0.0287}}$$ = 24.14 sec

According to Arrhenius equation,

$$k = A{e^{ - {{{E_a}} \over {ET}}}}$$

Taking natural log on both the sides we get,

ln k = ln A $${ - {{{E_a}} \over {ET}}}$$ ...........(1)

Comparing (1) with standard form of equation of line

y = mx + C

We get Slope, m = $${ - {{{E_a}} \over R}}$$

Hence, if ln k is plotted against 1/T, slope of the line will be $${ - {{{E_a}} \over R}}$$.