MP Board · Class 12 · Chemistry · Chemical KineticsThe thermal decomposition of $N2O5$ in gas phase proceeds according to the equation: $2N2O5(g) \rightarrow 4NO2(g) + O2(g)$. \nIf the initial concentration of $N2O5$ is $3.0 \times 10^{-2} \text{ mol L}^{-1}$ and it drops to $1.5 \times 10^{-2} \text{ mol L}^{-1}$ after 30 minutes, calculate the average rate of the reaction in terms of seconds and minutes. Also calculate the rate of production of $NO2$ during this time interval.

Step-by-Step Solution:

Given data:

• Initial concentration of $N_2O_5$, $[R]_1 = 3.0 \times 10^{-2} \text{ mol L}^{-1}$
• Final concentration of $N_2O_5$, $[R]_2 = 1.5 \times 10^{-2} \text{ mol L}^{-1}$
• Time interval, $\Delta t = 30 \text{ minutes} = 30 \times 60 = 1800 \text{ seconds}$

Step 1: Calculate the change in concentration of $N_2O_5$ ($\Delta [N_2O_5]$) $$\Delta [N_2O_5] = [R]_2 - [R]_1 = 1.5 \times 10^{-2} - 3.0 \times 10^{-2} = -1.5 \times 10^{-2} \text{ mol L}^{-1}$$

Step 2: Calculate the average rate of reaction in terms of minutes $$\text{Average Rate} = - \frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t}$$ $$\text{Average Rate} = - \frac{1}{2} \left( \frac{-1.5 \times 10^{-2} \text{ mol L}^{-1}}{30 \text{ min}} \right)$$ $$\text{Average Rate} = \frac{1.5 \times 10^{-2}}{60} = 2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}$$

Step 3: Calculate the average rate of reaction in terms of seconds $$\text{Average Rate in seconds} = \frac{- \frac{1}{2} \Delta [N_2O_5]}{\Delta t \text{ (in seconds)}}$$ $$\text{Average Rate} = - \frac{1}{2} \left( \frac{-1.5 \times 10^{-2}}{1800} \right) = \frac{1.5 \times 10^{-2}}{3600} = 4.17 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1}$$

Step 4: Calculate the rate of production of $NO_2$\nFrom the stoichiometry of the reaction, the rate expression is: $$\text{Rate} = - \frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = + \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t}$$\nTherefore, the rate of production of $NO_2$ is: $$\frac{\Delta [NO_2]}{\Delta t} = - 2 \times \frac{\Delta [N_2O_5]}{\Delta t}$$ $$\frac{\Delta [NO_2]}{\Delta t} = 2 \times (2.5 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}) = 5.0 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1}$$\nOr in terms of seconds: $$\frac{\Delta [NO_2]}{\Delta t} = 4 \times (4.17 \times 10^{-6}) = 1.67 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$$