Animated Solution for Chemistry - Chemical Kinetics: The decomposition reaction $2{\text{N}}_2{\text{O}}_5(\text{g})\stackrel{\Delta }{\to }2{\text{N}}_2{\text{O}}_4(\text{g})+{\text{O}}_2(\text{g})$ is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After $\text{Y}\times 10^3\text{ s}$, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is $5\times 10^{-4}{\text{ s}}^{-1}$, assuming ideal gas behavior, the value of Y is ______.

Imagine a closed steel cylinder acting as our reaction vessel. Inside, we have dinitrogen pentoxide ($N_2{O}_5$) gas ready to decompose. The problem explicitly states that the conditions are isothermal (constant temperature) and isochoric (constant volume). This is a crucial piece of information because, under constant volume and temperature, the partial pressure of any ideal gas is directly proportional to its number of moles ($P\propto n$).

Initially, the pressure gauge reads exactly $1\text{ atm}$. This is our starting point, $P_0=1\text{ atm}$.

To understand how the pressure evolves over time, we need to set up an ICE (Initial, Change, Equilibrium/Time t) table. Let's look at the balanced chemical equation:

For every $2$ moles of ${\text{N}}_2{\text{O}}_5$ that decompose, $2$ moles of ${\text{N}}_2{\text{O}}_4$ and $1$ mole of ${\text{O}}_2$ are formed. Translating this to partial pressures, if the pressure of ${\text{N}}_2{\text{O}}_5$ drops by $2P$, the products gain $2P$ and $P$ respectively.

At time $t$, the partial pressures are: - $P_{{\text{N}}_2{\text{O}}_5}=1-2P$ - $P_{{\text{N}}_2{\text{O}}_4}=2P$ - $P_{{\text{O}}_2}=P$

According to Dalton's Law of Partial Pressures, the total pressure inside the cylinder is the sum of these individual pressures:

We are given that the total pressure at time $t$ is $1.45\text{ atm}$. Equating this to our expression:

Now, we can find the remaining partial pressure of our reactant, ${\text{N}}_2{\text{O}}_5$, at time $t$:

Here is where many students fall into a trap. Look at the unit of the rate constant: $5\times 10^{-4}{\text{ s}}^{-1}$. The unit ${\text{s}}^{-1}$ screams first-order kinetics!

However, we must be extremely careful with the stoichiometry. The standard rate of reaction is defined as:

Rearranging this gives the rate of disappearance of ${\text{N}}_2{\text{O}}_5$:

Because the coefficient of ${\text{N}}_2{\text{O}}_5$ is $2$, the effective rate constant in our integrated rate law becomes $2k$. The integrated first-order rate law is:

Let's substitute our known values into the master equation:

The left side simplifies beautifully: $2\times 5\times 10^{-4}=10\times 10^{-4}=10^{-3}$. On the right side, $\frac{1}{0.1}=10$, and we know that $\log (10)=1$.

Moving the power of ten to the right side, we get:

The problem states that the time is $\text{Y}\times 10^3\text{ s}$. Comparing this with our result, we find that $\text{Y}=2.303$. Rounding to two decimal places as per standard numerical entry formats, our final answer is 2.30.