b
\({T_2} = T\) (say),  \(T = {25\,^o}C = 298\,K\),

\({E_a} = 104.4\,{\rm{kJ}}\,mo{l^{ - 1}} = 104.4 \times {10^3}\,{\rm{J}}\,mo{l^{ - 1}}\)

\({K_1} = 3 \times {10^{ - 4}},\,\,{K_2} = \)?,

\(\log \frac{{{K_2}}}{{{K_1}}} = \frac{{{E_a}}}{{2.303R}}\left[ {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right]\)

\(\log \frac{{{K_2}}}{{3 \times {{10}^{ - 4}}}} = \frac{{104.4 \times {{10}^3}{\rm{J}}\,\,mo{l^{ - 1}}}}{{2.303 \times (8.314\,\,{\rm{J}}\,\,{k^{ - 1}}\,\,mo{l^{ - 1}})}}\)

\(\left[ {\frac{1}{{298\,K}} - \frac{1}{T}} \right]\,\,{\rm{As}}\,\,{\rm{T}} \to \infty {\rm{,}}\frac{{\rm{1}}}{{\rm{T}}} \to 0\)

\(\therefore \,\,\log \frac{{{K_2}}}{{3 \times {{10}^{ - 4}}}} = \frac{{104.4 \times {{10}^3}\,\,{\rm{J}}\,\,mo{l^{ - 1}}}}{{2.303 \times 8.314 \times 298}}\)

\(\log \frac{{{K_2}}}{{3 \times {{10}^{ - 4}}}} = 18.297,\,\,\frac{{{K_2}}}{{3 \times {{10}^{ - 4}}}}\) \( = 1.98 \times {10^{18}}\) or

\({K_2} = (1.98 \times {10^{18}}) \times (3 \times {10^{ - 4}}) = 6 \times {10^{14}}\,{s^{ - 1}}\)