Chemistry: The reaction 2N2O5 (g) → 2N2O4 (g) + O2 (g) has a reaction rate that is dependent only on the concentration of N2O5 and at a certain temperature has a rate constant k of 0.0168 s-1. If 2.50 moles of N2O5 were placed in a 5.00 liter container at that temperature, how many moles of N2O5 would remain after 1.00 min?

The reaction 2N2O5 (g) → 2N2O4 (g) + O2 (g) has

june1198

23.01.2022

0.910 mol

Explanation:

Let's consider the following reaction.

2 N₂O₅(g) → 2 N₂O₄(g) + O₂(g)

This reaction is of first order with respect to N₂O₅, with a rate constant k = 0.0168 s⁻¹.

We can calculate the concentration at a certain time using the following expression.

$[N_2O_5] = [N_2O_5]_0 \times e^{-k \times t}$

where,

: initial concentrationk: rate constantt: time

The initial concentration of N₂O₅ is:

2.50 mol / 5.00 L = 0.500 M

The concentration of N₂O₅ after 1.00 min (60 s) is:

$[N_2O_5] = 0.500 M \times e^{-0.0168 s^{-1} \times 60s} = 0.182 M$

The moles of N₂O₅ after 1.00 min are:

$5.00 L \times \frac{0.182mol}{L} =0.910 mol$

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