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Q3.15(i):

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:

(i) Plot \([N_{2}O_{5}]\) against \(t\).

Solution :

Step 1: Data Tabulation and Variable Assignment

To analyze the decomposition of $N_{2}O_{5}$, we represent the concentration $[N_{2}O_{5}]$ as the dependent variable ($y$-axis) and time $t$ as the independent variable ($x$-axis) based on the provided experimental set:

Time $t$ (s) $[N_{2}O_{5}]$ (mol L$^{-1}$)
0 1.63
100 1.36
200 1.14
300 0.93
400 0.78
500 0.64
600 0.53
700 0.43

Step 2: Graphical Representation Protocol

Plotting $[N_{2}O_{5}]$ vs $t$ yields a downward-sloping curve, characteristic of a first-order kinetic process [Per Rate Law: $Rate = k[N_{2}O_{5}]$]. As time progresses, the concentration decreases non-linearly, indicating that the reaction rate slows down as reactants are consumed.

Time (t) in seconds [N₂O₅] (mol L⁻¹)

Step 3: Analytical Interpretation

The resulting graph demonstrates that the concentration of $N_{2}O_{5}$ decays exponentially with respect to time. By applying the First-Order Integrated Rate Equation: $\ln[N_{2}O_{5}]_t = -kt + \ln[N_{2}O_{5}]_0$ [By Fundamental Law of Chemical Kinetics], the slope of the tangent at any point $t$ provides the instantaneous rate of the reaction at that specific time interval.

Final Solution: The plot of $[N_{2}O_{5}]$ versus $t$ results in an exponential decay curve, confirming that the concentration of the reactant decreases as the reaction proceeds towards equilibrium.


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