91Ó°ÊÓ

First, we need to convert the given temperatures 25°C and 125°C to Kelvin. To convert Celsius to Kelvin, use this equation: Temperature(K) = Temperature(°C) + 273.15 T1 = 25 + 273.15 = 298.15 K T2 = 125 + 273.15 = 398.15 K

Let's set up the ratio of the rate constants with the catalyst (kc) to without the catalyst (ku) for each temperature: T1: \(\frac{k_{c1}}{k_{u1}} = \frac{Ae^{\frac{-55}{8.314\times 298.15}}}{Ae^{\frac{-95}{8.314\times 298.15}}}\) T2: \(\frac{k_{c2}}{k_{u2}} = \frac{Ae^{\frac{-55}{8.314\times 398.15}}}{Ae^{\frac{-95}{8.314\times 398.15}}}\)

We can cancel the collision factor A since it remains the same for both reactions. This leaves us with equations involving only the activation energies and temperatures: T1: \(\frac{k_{c1}}{k_{u1}} = \frac{e^{\frac{-55}{8.314\times 298.15}}}{e^{\frac{-95}{8.314\times 298.15}}}\) T2: \(\frac{k_{c2}}{k_{u2}} = \frac{e^{\frac{-55}{8.314\times 398.15}}}{e^{\frac{-95}{8.314\times 398.15}}}\) Now we can solve for the fractions: T1: \(\frac{k_{c1}}{k_{u1}}\approx 9.82\) T2: \(\frac{k_{c2}}{k_{u2}}\approx 20.64\) These values tell us how much faster the reaction is with the catalyst compared to without. At 25°C (298.15 K), the catalyst increases the rate of the reaction by a factor of approximately 9.82. At 125°C (398.15 K), the catalyst increases the rate of the reaction by a factor of approximately 20.64.