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Q3.15(iv):

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: 

(iv) What is the rate law ?

Solution :

Step 1: Analyzing the Reaction Order

The general rate law for the decomposition reaction $2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}$ is expressed as:

$\text{Rate} = k[N_{2}O_{5}]^n$

[By definition of the rate law for a single reactant system].

Step 2: Calculating Rate of Reaction at Intervals

The rate of reaction is given by $-\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt}$. Using the provided data points, we calculate the average rate over specific intervals ($\Delta t$):

Time ($t$) / s $[N_{2}O_{5}] / \text{mol L}^{-1}$ Average Rate ($\frac{-\Delta [N_{2}O_{5}]}{2 \Delta t}$) / $\text{mol L}^{-1} \text{s}^{-1}$
0 1.63 -
400 1.36 $\frac{-(1.36 - 1.63)}{2 \times 400} = 3.375 \times 10^{-4}$
800 1.14 $\frac{-(1.14 - 1.36)}{2 \times 400} = 2.75 \times 10^{-4}$

Step 3: Determining the Rate Constant ($k$) for First Order

For a first-order reaction ($n=1$), the integrated rate equation is:

$k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$

[Per the Integrated Rate Law for First-Order Kinetics].

  • At $t = 400\text{s}$: $k = \frac{2.303}{400} \log \frac{1.63}{1.36} = 0.00575 \times 0.18 = 1.03 \times 10^{-3} \text{s}^{-1}$
  • At $t = 800\text{s}$: $k = \frac{2.303}{800} \log \frac{1.63}{1.14} = 0.00287 \times 0.35 = 1.00 \times 10^{-3} \text{s}^{-1}$

Step 4: Graphical Representation of First Order Kinetics

Time (t) ln[N₂O₅] ln[N₂O₅] = -kt + ln[N₂O₅]₀

Step 5: Establishing the Rate Law

Since the value of $k$ remains approximately constant at $1.02 \times 10^{-3} \text{s}^{-1}$, the reaction is confirmed to be of the first order with respect to $N_{2}O_{5}$.

Final Solution: The rate law is $\text{Rate} = k[N_{2}O_{5}]^1$ where $k \approx 1.02 \times 10^{-3} \text{s}^{-1}$.


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