The decomposition of $A$ into product has a value of $k$ as $4.5 \times 10^{3} \, s^{-1}$ at $10^{\circ} C$ and an energy of activation of $60 \, kJ \, mol^{-1}$. At what temperature would $k$ be $1.5 \times 10^{4} \, s^{-1}$?

(D) From the Arrhenius equation,we have:
$\log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 R} \left( \frac{T_{2} - T_{1}}{T_{1} T_{2}} \right)$
Given:
$k_{1} = 4.5 \times 10^{3} \, s^{-1}$,$T_{1} = 283 \, K$,$k_{2} = 1.5 \times 10^{4} \, s^{-1}$,$E_{a} = 60,000 \, J \, mol^{-1}$,$R = 8.314 \, J \, K^{-1} \, mol^{-1}$.
Substituting the values:
$\log \left( \frac{1.5 \times 10^{4}}{4.5 \times 10^{3}} \right) = \frac{60000}{2.303 \times 8.314} \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$\log(3.333) = 3133.6 \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$0.5229 = 3133.6 \left( \frac{T_{2} - 283}{283 T_{2}} \right)$
$0.0472 = \frac{T_{2} - 283}{T_{2}}$
$0.0472 T_{2} = T_{2} - 283$
$0.9528 T_{2} = 283$
$T_{2} = 297 \, K = 24^{\circ} C$.