If the initial concentration of N2O5 is 0.117 M, the concentration of N2O5 will be 3.12×10-2 M after ______s have passed.

Question

The gas phase decomposition of dinitrogen pentoxide at 335 K

N₂O₅(g)2 NO₂(g) + ½ O₂(g)

is first order in N₂O₅ with a rate constant of 4.70×10^-3s^-1.

If the initial concentration of N₂O₅ is 0.117 M, the concentration of N₂O₅ will be 3.12×10^-2 M after ______s have passed.

Answer 1

For a 1st order reaction

ln [A] = -kt ln [A]_o

ln [3.12x10^-2 M] = -(4.70x10^-3 s^-1)(t) ln 0.117 M

-1.32 / 4.70x10^-3 = t

t = 281 s

Answer 2

From the data, the given reaction is of the first order.

The integrated rate law of a first-order reaction is as follows:

ln [N₂O₅] / [ N₂O₅]₀ = -kt

Let the time be 'Q' seconds

Therefore, the concentration of N₂O₅ after X sec is calculated as follows:

ln 3.12×10^-2 M / 0.117 M = - 4.70×10^-3s^-1 x Q s

Solving Q from the above, we get 281 seconds

Hence, If the initial concentration of N₂O₅ is 0.117 M, the concentration of N₂O₅ will be 3.12×10^-2 M after 281 s have passed.

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