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Q3.15(ii):

The experimental data for decomposition of \(N_{2}O_{5}\) [\(2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}\)] in gas phase at 318K are given below:

(ii) Find the half-life period for the reaction.

Solution :

Step 1: Determining the Order of Reaction and Rate Constant

Based on the provided experimental data for the decomposition of $N_{2}O_{5}$ at $318\text{ K}$, the reaction follows first-order kinetics. This is confirmed by the linear relationship between $\ln[N_{2}O_{5}]$ and time, or by checking the consistency of the rate constant $k$ using the first-order integrated rate law: $k = \frac{2.303}{t} \log \frac{[A]_{0}}{[A]_{t}}$ [Per the First-Order Integrated Rate Law].

Using the data points at $t = 0$ ($[A]_{0} = 1.63 \times 10^{-2} \text{ M}$) and $t = 184 \text{ s}$ ($[A]_{t} = 1.36 \times 10^{-2} \text{ M}$):

$k = \frac{2.303}{184} \log \frac{1.63 \times 10^{-2}}{1.36 \times 10^{-2}}$ $k = \frac{2.303}{184} \log(1.1985)$ $k = \frac{2.303}{184} \times 0.0786 \approx 9.84 \times 10^{-4} \text{ s}^{-1}$

Step 2: Deriving the Half-Life Period

For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration and is defined as the time taken for the concentration of the reactant to reduce to half of its initial value. The relationship is derived from the integrated rate law where $[A]_{t} = \frac{[A]_{0}}{2}$ at $t = t_{1/2}$:

$k = \frac{2.303}{t_{1/2}} \log \frac{[A]_{0}}{[A]_{0}/2}$ $k = \frac{2.303}{t_{1/2}} \log(2)$ $t_{1/2} = \frac{0.693}{k}$ [Per the definition of First-Order Half-life].

Step 3: Numerical Calculation

Substituting the calculated value of $k$ into the half-life formula:

$t_{1/2} = \frac{0.693}{9.84 \times 10^{-4} \text{ s}^{-1}}$ $t_{1/2} \approx 704.27 \text{ s}$

Visual Representation of Decomposition

Concentration vs Time (First Order) Time (s) [N₂O₅] t₁/₂ ≈ 704s

Final Solution: The half-life period ($t_{1/2}$) for the decomposition of $N_{2}O_{5}$ at 318 K is approximately 704.27 seconds.


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