Explanation

(a)
To be solved with the help of formula,
$\ \log\frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303\ R}\left\lbrack \frac{T_{2} - T_{1}}{T_{1}T_{2}} \right\rbrack$
$\ T_{1} = 273 + 27 = 300\ K$
$\ T_{2} = 273 + 67 = 340\ K$
$\log\frac{6.9 \times 10^{- 3}}{3.45 \times 10^{- 5}} = \frac{E_{a}}{2.303 \times 8.31}\left\lbrack \frac{340 - 300}{340 \times 300} \right\rbrack$
$\log 200 = \frac{E_{a}}{19.1379} \times \frac{40}{102000}$
$2.3010 = \frac{E_{a}}{19.14} \times \frac{4}{10200},$
$\ E_{a} = \frac{19.14 \times 10200 \times 2.3010}{4}$
$= 112304.907\ J = 112.3\ kJ$