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 $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: