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 $${ - {{{E_a}} \over R}}$$.

For zero order reaction unit of rate constant is mole per second.

$$ \therefore $$ For zero order

x = kt

$$ \Rightarrow $$ x = 0.6 × 10^–3 × 20 × 60 = 0.72 M

For a first order reaction,

t_75% = 2 $$ \times $$ t_50%

$$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303R}}\left( {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right)$$

Initial temperature, T₁ = 20 + 273 = 293 K

Final temperature, T₂ = 35 + 273 = 308 K

R = 8.314 JK^–1 mol^–1

As rate becomes double on raising temperature

$$ \therefore $$ r₂ = 2r₁

As rate constant, k $$\infty $$ r

k₂ = 2k₁

$$ \therefore $$ $$\log 2 = {{{E_a}} \over {2.303 \times 8.314}}\left( {{1 \over {293}} - {1 \over {308}}} \right)$$

$$ \Rightarrow $$ $$0.301 = {{{E_a}} \over {19.147}} \times {{15} \over {293 \times 308}}$$

$$ \Rightarrow $$ E_$$a$$ = 34673 J mol^–1 = 34.7 kJ mol^–1