First Order Kinetics and Pressure Calculation for Dinitrogen Pentoxide Decomposition

To find the value of $Y$, we analyze the reaction stoichiometry and first-order chemical kinetics.

1. Stoichiometric Analysis

The given decomposition reaction is: $2\text{N}_2\text{O}_5(g) \xrightarrow{\Delta} 2\text{N}_2\text{O}_4(g) + \text{O}_2(g)$

Since the reaction takes place in a closed cylinder under isothermal and isochoric conditions, the pressure of each gas is directly proportional to its molar concentration ($P \propto n$ via the ideal gas equation $P = \frac{nRT}{V}$).

Let the initial pressure of $\text{N}_2\text{O}_5$ be $P_0 = 1\text{ atm}$.

At time $t$:

• Let the decrease in pressure of $\text{N}_2\text{O}_5$ be $2x$.
• The partial pressure of $\text{N}_2\text{O}_4$ produced is $2x$.
• The partial pressure of $\text{O}_2$ produced is $x$.

The partial pressures at time $t$ are: $P_{\text{N}_2\text{O}_5} = P_0 - 2x$ $P_{\text{N}_2\text{O}_4} = 2x$ $P_{\text{O}_2} = x$

The total pressure inside the cylinder at time $t$ is: $P_{\text{total}} = P_{\text{N}_2\text{O}_5} + P_{\text{N}_2\text{O}_4} + P_{\text{O}_2} = (P_0 - 2x) + 2x + x = P_0 + x$

Given that $P_{\text{total}} = 1.45\text{ atm}$ and $P_0 = 1\text{ atm}$: $1.45 = 1 + x \implies x = 0.45\text{ atm}$

Now, we calculate the remaining partial pressure of $\text{N}_2\text{O}_5$ at time $t$: $P_{\text{N}_2\text{O}_5} = P_0 - 2x = 1 - 2(0.45) = 1 - 0.90 = 0.10\text{ atm}$

---

2. Kinetic Calculation

By IUPAC convention, the rate of reaction for $2\text{N}_2\text{O}_5(g) \to 2\text{N}_2\text{O}_4(g) + \text{O}_2(g)$ is defined as: $\text{Rate} = -\frac{1}{2}\frac{d[\text{N}_2\text{O}_5]}{dt} = k[\text{N}_2\text{O}_5]$

In terms of partial pressure: $-\frac{1}{2}\frac{dP_{\text{N}_2\text{O}_5}}{dt} = k P_{\text{N}_2\text{O}_5}$

Rewriting in terms of the rate of disappearance of $\text{N}_2\text{O}_5$: $-\frac{dP_{\text{N}_2\text{O}_5}}{dt} = 2k P_{\text{N}_2\text{O}_5}$

Integrating this first-order rate law from $t = 0$ (where $P = P_0$) to time $t$ (where $P = P_{\text{N}_2\text{O}_5}$): $\int_{P_0}^{P_{\text{N}_2\text{O}_5}} \frac{dP_{\text{N}_2\text{O}_5}}{P_{\text{N}_2\text{O}_5}} = -2k \int_0^t dt$

$\ln\left(\frac{P_0}{P_{\text{N}_2\text{O}_5}}\right) = 2kt$

Solving for time $t$: $t = \frac{1}{2k}\ln\left(\frac{P_0}{P_{\text{N}_2\text{O}_5}}\right)$

Substitute the given values:

• $k = 5 \times 10^{-4}\text{ s}^{-1}$
• $P_0 = 1\text{ atm}$
• $P_{\text{N}_2\text{O}_5} = 0.10\text{ atm}$

$t = \frac{1}{2 \times (5 \times 10^{-4}\text{ s}^{-1})} \ln\left(\frac{1}{0.10}\right)$ $t = \frac{1}{10^{-3}} \ln(10) = 10^3 \times 2.3026\text{ s} \approx 2.30 \times 10^3\text{ s}$

We are given that $t = Y \times 10^3\text{ s}$. Comparing the two expressions: $Y \times 10^3 = 2.30 \times 10^3 \implies Y = 2.30$

(Note: If the rate constant is alternatively defined with respect to the reactant disappearance as $-\frac{dP_{\text{N}_2\text{O}_5}}{dt} = k P_{\text{N}_2\text{O}_5}$, then $t = \frac{\ln(10)}{k} = \frac{2.303}{5 \times 10^{-4}} = 4.60 \times 10^3\text{ s}$, which gives $Y = 4.60$. Standard convention designates $Y = 2.30$.)

Thus, the value of $Y$ is 2.30 (or 2.3).