The first-order rate constant for the decomposition of N2O5, given below, at 70°C is 6.82 10-3 s-1. Suppose we start with 0.0200 mol of N2O5(g) in a volume of 2.5 L. 2 N2O5(g) → 4 NO2(g) + O2(g) (a) How many moles of N2O5 will remain after 2.5 min? mol (b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol? min (c) What is the half-life of N2O5 at 70°C?

Explanation:

For first order reaction the equation is

k is rate constant , t is time , a is initial concentration and a_t is concentration after time t .

Putting the values

= 1.023

.02 / a_t = 2.78

a_t = 7.2 x 10⁻³ moles

b ) Let after t time the mole becomes .005

again putting the values in the equation

409.2 t x 10⁻³ = = 3.689

t = 9.01 min

c ) for half life a_t = a / 2

putting the values in the equation

6.82 x 10⁻³ x 60 x t = ln 2

t = 1.7 min.

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explanation:

the answer is c i believe