From Arrhenius equation, $$\log k = \log A - {{{E_a}} \over {2.303RT}}$$

or, $$\log {{{k_2}} \over {{k_1}}} = {{{E_a}} \over {2.303}}\left[ {{1 \over {{T_1}}} - {1 \over {{T_2}}}} \right]$$

$$\therefore$$ $$\log {{{k_2}} \over {{k_1}}} = {{70} \over {2.303 \times 8.314}}\left[ {{1 \over {298}} - {1 \over {313}}} \right]$$ or, $${{{k_2}} \over {{k_1}}} = 3.872$$ ...... [1]

For a first order reaction, $$k = {{2.303} \over t}\log {a \over {a - x}}$$

Given x = 0.25 a,

$$\therefore$$ a $$-$$ x = (a $$-$$ 0.25 a) = 0.75 a at t = 20 min

$$\therefore$$ $${k_1} = {{2.303} \over {20}}\log {a \over {0.75\,a}} = 0.014386$$ min^$$-$$1 ...... [2]

Substituting for k₁ in equation 1, we have

$${k_2} = 3.872 \times 0.014386 = 0.05571$$ min^$$-$$1

For a first order reaction, $${k_2} = {{2.303} \over t}\log {a \over {a - x}}$$

or, $$0.05571 = {{2.303} \over {20}}\log {a \over {a - x}}$$ [x = decrease in concentration of reactant]

or, $$\log {a \over {a - x}} = {{0.05571 \times 20} \over {2.303}}$$

or, $$\log {a \over {a - x}} = 0.48381$$

or, x = 0.6717 a

Hence, percentage decomposition $$ = {{0.6717\,a} \over a} \times 100 = 67.17\% $$