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The decomposition of N(2)O(5) in CCl(4) ...

The decomposition of `N_(2)O_(5)` in `CCl_(4)` solution follows the first order rate law. The concentrations of `N_(2)O_(5)` measured at different intervals are given below :
`{:("Time in seconds (t)",,,0,,,80,,,160,,,410,,,600,,,1130,,,1720),(["N"_(2)"O"_(5)]"mol"//"L",,,5.5,,,5.0,,,4.8,,,4.0,,,3.4,,,2.4,,,1.6):}`
Calculate the rate constant at t = 410 s and t = 1130 s. What do these results show ?

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AI Generated Solution

To solve the problem of calculating the rate constant for the decomposition of \(N_2O_5\) at two different times (410 seconds and 1130 seconds), we will use the first-order rate law equation. The equation for the rate constant \(k\) in a first-order reaction is given by: \[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \] where: - \(t\) is the time, ...
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