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
Step 1: Write the rate constant expressions using the Arrhenius equation.
\[ K_1 = A_1 e^{-E_{a1}/RT}, \quad K_2 = A_2 e^{-E_{a2}/RT}, \quad K_3 = A_3 e^{-E_{a3}/RT} \]
\[ K = \frac{K_1 K_2}{K_3} = \frac{A_1 A_2}{A_3} \, e^{-(E_{a1} + E_{a2} - E_{a3})/RT} \]
From the exponent, \[ E_a = E_{a1} + E_{a2} - E_{a3} \]
\[ E_a = 40 + 50 - 60 = 30\,\text{kJ/mol} \]
\[ \boxed{E_a = 30\,\text{kJ/mol}} \]
Consider the following compounds. Arrange these compounds in a n increasing order of reactivity with nitrating mixture. The correct order is : 
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 