For the overall rate constant:
\[ K = \frac{K_1 \cdot K_2}{K_3} = \frac{A_1 \cdot A_2}{A_3} \cdot e^{\left(\frac{E_{a1} + E_{a2} - E_{a3}}{RT}\right)} \]
Therefore,
\[ K = \frac{A \cdot e^{-E_a/RT}}{A_3} = \frac{A_1 A_2}{A_3} \cdot e^{\left(\frac{E_{a1} + E_{a2} - E_{a3}}{RT}\right)} \]
Given:
\[ E_a = E_{a1} + E_{a2} - E_{a3} = 40 + 50 - 60 = 30 \, \text{kJ/mol} \]
So, the correct answer is: 30
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32