For a secondary bromide in ethanol, elimination has a higher activation enthalpy than substitution…

Step 1. Crossover. $\Delta\Delta G^\ddagger = 0$ gives $T = \dfrac{10\,000\ \text{J/mol}}{30.0\ \text{J/(mol K)}} = 333\ \text{K}$.

Step 2. At 298 K. $\Delta\Delta G^\ddagger = 10\,000 - 298 \times 30.0 = 1060\ \text{J/mol}$. Then $r = e^{-1060\ \text{J/mol} / (8.314\ \text{J/(mol K)} \times 298\ \text{K})} = e^{-0.428} = 0.652$. The share is $\dfrac{0.652}{1.652} \times 100 = 39.5\ \%$.

Step 3. At 350 K. $\Delta\Delta G^\ddagger = 10\,000 - 350 \times 30.0 = -500\ \text{J/mol}$. Then $r = e^{500\ \text{J/mol} / (8.314\ \text{J/(mol K)} \times 350\ \text{K})} = e^{0.172} = 1.19$. The share is $\dfrac{1.19}{2.19} \times 100 = 54.3\ \%$.