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Question
The experimental data for decomposition of N2O5
\[\ce{2N2O5 -> 4NO2 + O2}\]
in gas phase at 318 K are given below:
| t/s | 0 | 400 | 800 | 1200 | 1600 | 2000 | 2400 | 2800 | 3200 |
| 102 × [N2O5]/mol L−1 | 1.63 | 1.36 | 1.14 | 0.93 | 0.78 | 0.64 | 0.53 | 0.43 | 0.35 |
- Plot [N2O5] against t.
- Find the half-life period for the reaction.
- Draw a graph between log [N2O5] and t.
- What is the rate law?
- Calculate the rate constant.
- Calculate the half-life period from k and compare it with (ii).
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Solution
| t/s | [N2O5] × 102/mol L−1 | log [N2O5] |
| 0 | 1.63 | −1.79 |
| 400 | 1.36 | −1.87 |
| 800 | 1.14 | −1.94 |
| 1200 | 0.93 | −2.03 |
| 1600 | 0.78 | −2.11 |
| 2000 | 0.64 | −2.19 |
| 2400 | 0.53 | −2.28 |
| 2800 | 0.43 | −2.37 |
| 3200 | 0.35 | −2.46 |
i.
ii. Initial concentration of N2O5 = 1.63 × 10−2 M
Half of this concentration = 0.815 × 10−2 M
The time corresponding to M concentration = 1440 s
Hence, t1/2 = 1440 s
iii.
iv. The given reaction is of the first order as the plot, log [N2O5] v/s t, is a straight line. Hence, it is a reaction of first order, i.e., the rate law is:
Rate = k[N2O5]
v. For first order reaction,
log R = `- k/(2.303 t) + log R_0`
Therefore, the slope of the graph drawn between log R and t will be `(-k)/2.303`.
∴ The slope of the line = `-k/2.303 = (y_2 - y_1)/(t_2 - t_1)`
`-k/2.303 = (-2.46 - (-1.79))/(3200 - 0)`
`-k/2.303 = -0.67/3200`
`k/2.303 = 0.67/3200`
k = `0.67/3200 xx 2.303`
∴ k = 4.82 × 10−4 s−1
vi. Half-life is given by,
t1/2 = `0.639/k`
= `0.693/(4.82 xx 10^-4)`
= 1.438 × 103 s
= 1438 s
This value, 1438 s, is very close to the value that was obtained from the graph.
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