contestada

The first-order rate constant for the decomposition of N2O5:
2N2O5(g) -------> 4NO2(g)+O2(g)
at 70°C is 6.82 x 10^−3 s−1. Suppose we start with 2.10 x 10^−2 mol of N2O5(g) in a volume of 1.8 L.
(A) How many moles of N2O5 will remain after 7.0 min ? Express the amount in moles to two significant digits.

Respuesta :

princessesther2011

Answer:

0.0055 mol of N2O5 will remay after 7 min.

Explanation:

The reaction follows a first-order.

Let the concentration of N2O5 after 7 min be y

Rate = Ky = change in concentration of N2O5/time

K is rate constant = 6.82×10^-3 s^-1

Initial concentration of N2O5 = number of moles/volume = 2.1×10^-2/1.8 = 0.0117 M

Change in concentration = 0.0117 - y

Time = 7 min = 7×60 = 420 s

6.82×10^-3y = 0.0117 - y/420

0.0117 - y = 420×6.82×10^-3y

0.0117 - y = 2.8644y

0.0117 = 2.8644y + y

0.0117 = 3.8644y

y = 0.0117/3.8644 = 0.00303 M

Number of moles of N2O5 left = y × volume = 0.00303 × 1.8 = 0.0055 mol (to 2 significant digits)

Otras preguntas