The decomposition of N₂O₅ in solution in carbon tetrachloride proceeds via the reaction 2N₂O₅ (soln) → 4NO₂ (soln) + O₂ (soln)
The reaction is first order and has a rate constant of 4.82 × 10^-3s^-1 at 64°C. If the reaction is initiated with 0.058 mol in a 1.00-L vessel, how many moles remain after 151 s?

Question

Quizplus · Accepted Answer

The given reaction is a first-order reaction. The integrated rate equation for a first-order reaction is:

ln([A]t/[A]0) = -kt

where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is time.

We are given [A]0 = 0.058 mol, k = 4.82 × 10^-3 s^-1, and t = 151 s. We are asked to find [A]t.

Plugging in the values, we get:

ln([A]t/0.058) = -4.82 × 10^-3 × 151
ln([A]t/0.058) = -0.3265
[A]t/0.058 = e^-0.3265
[A]t = 0.028 mol

Therefore, the amount of A remaining after 151 s is 0.028 moles.

The answer is (C) 0.028.

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The Decomposition of N₂O₅ in Solution in Carbon Tetrachloride Proceeds