Step-by-step Derivations:

1. Analyze the Reaction and Given Data

The decomposition reaction is: $2{N_2}{O_5}(g)\xrightarrow{\Delta} 2{N_2}{O_4}(g) + {O_2}(g)$

• The reaction is conducted under isothermal (constant temperature) and isochoric (constant volume) conditions.
• For an ideal gas under these conditions, pressure is directly proportional to the number of moles ($P \propto n$). Therefore, we can use partial pressures in place of concentrations.
• The unit of the rate constant ($s^{-1}$) indicates that this is a first-order reaction.
• Initial pressure of $N_2O_5$, $P_0 = 1$ atm.
• Total pressure after time $t$, $P_{total} = 1.45$ atm.
• Rate constant of the reaction, $k = 5 \times 10^{-4} s^{-1}$.
• Time elapsed, $t = Y \times 10^3$ s.

2. Relate Total Pressure to the Partial Pressure of the Reactant

Let's set up a table to track the partial pressures:

The total pressure at time $t$ is the sum of the partial pressures: $P_{total} = P_{N_2O_5} + P_{N_2O_4} + P_{O_2}$ $P_{total} = (1 - 2x) + 2x + x = 1 + x$

We are given that at time $t$, $P_{total} = 1.45$ atm. $1.45 = 1 + x$ $x = 1.45 - 1 = 0.45 \text{ atm}$

Now, we can find the partial pressure of $N_2O_5$ at time $t$, let's call it $P_t$: $P_t = P_{N_2O_5} \text{ at time t} = 1 - 2x = 1 - 2(0.45) = 1 - 0.90 = 0.10 \text{ atm}$

3. Apply the Integrated Rate Law

For a general first-order reaction $aA \to \text{products}$, the rate of reaction ($r$) is defined as $r = -\frac{1}{a}\frac{d[A]}{dt}$. If the rate law is given by $r = k[A]$, then: $-\frac{1}{a}\frac{d[A]}{dt} = k[A] \implies \frac{d[A]}{dt} = -ak[A]$

The integrated rate law is therefore: $\ln\frac{[A]_0}{[A]_t} = akt$

For our reaction, $2N_2O_5 \to \text{products}$, the stoichiometric coefficient $a = 2$. The rate constant given, $k$, is for the reaction rate. Thus, the integrated rate law in terms of partial pressures is: $\ln\left(\frac{P_0}{P_t}\right) = 2kt$

4. Calculate the Time (t)

Substitute the known values into the integrated rate law:

• $P_0 = 1$ atm
• $P_t = 0.10$ atm
• $k = 5 \times 10^{-4} s^{-1}$

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

Using the value $\ln(10) \approx 2.303$: $2.303 = 10^{-3} \times t$ $t = \frac{2.303}{10^{-3}} = 2.303 \times 10^3 \text{ s}$

5. Determine the Value of Y

The problem states that the time is $Y \times 10^3$ s. By comparing this with our calculated time: $Y \times 10^3 = 2.303 \times 10^3$ $Y = 2.303$

Rounding to one decimal place as suggested by the stored answer format, we get $Y = 2.3$.