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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?

Respuesta :

dsdrajlin

Answer:

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.

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

where,

  • [tex][N_2O_5]_0[/tex]: initial concentration
  • k: rate constant
  • t: 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:

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

The moles of N₂O₅ after 1.00 min are:

[tex]5.00 L \times \frac{0.182mol}{L} =0.910 mol[/tex]

pstnonsonjoku

The number of moles of N2O5 after 1 minute is 0.90 moles.

We must first obtain the initial concentration of the solution, using the information;

Volume of the system = 5.00 liters

Initial number of moles present = 2.50 moles

Initial concentration = 2.50 moles/ 5.00 liters = 0.50 M

Since the rate of reaction depends only on the concentration of N2O5, the reaction is first order.

Hence;

lnA = lnAo - kt

A = concentration at time t

Ao = Initial concentration

k = rate constant

t = time taken

lnA = ln(0.50 M) - (0.0168 s-1 × 60 s)

lnA = -0.693 - 1.008

A = e^(-1.701)

A= 0.18 M

Number of moles = Concentration × volume

Since the volume of the solution did not change;

Number of moles = 0.18 M × 5.00 L = 0.90 moles

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