Catalysts lower activation energies equally for forward and backward reactions without altering ∆H
Step 1: Activation energy for the uncatalyzed backward reaction:
\( E_a(\text{backward}) = E_a(\text{forward}) - \Delta H \)
\( E_a(\text{backward}) = 300 - 20 = 280 \, \text{kJ/mol} \)
Step 2: Using the given temperatures and equal rates, calculate \( E_a(\text{forward, catalyzed}) \):
\( \frac{E_a(\text{forward, catalyzed})}{E_a(\text{forward, uncatalyzed})} = \frac{T_c}{T_u} \)
\( \frac{E_a(\text{forward, catalyzed})}{300} = \frac{300}{600} \)
\( E_a(\text{forward, catalyzed}) = 150 \, \text{kJ/mol} \)
Step 3: Calculate \( E_a(\text{backward, catalyzed}) \):
\( E_a(\text{backward, catalyzed}) = E_a(\text{forward, catalyzed}) - \Delta H \)
\( E_a(\text{backward, catalyzed}) = 150 - 20 = 130 \, \text{kJ/mol} \)
Let one focus of the hyperbola $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $ be at $ (\sqrt{10}, 0) $, and the corresponding directrix be $ x = \frac{\sqrt{10}}{2} $. If $ e $ and $ l $ are the eccentricity and the latus rectum respectively, then $ 9(e^2 + l) $ is equal to:
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is: