Question

The first-order rate constant for the decomposition of N2O5,

2N2O5(g)→4NO2(g)+O2(g)

at 70∘C is 6.82×10−3 s−1. Suppose we start with 2.30×10⁻² mol of N2O5(g) in a volume of 1.5 L .

a. How many moles of N2O5 will remain after 6 min ?

b. How many minutes will it take for the quantity of N2O5 to drop to 1.9×10⁻² mol ?

c. What is the half-life of N2O5 at 70∘C?

Answer 1

Given reaction is,

2N₂O₅ (g) → 4NO₂ (g) + O₂ (g)

The rate constant, k = 6.82 * 10^-3

The no. of moles of N₂O₅ = 2.30 * 10^-2

Volume, V = 1.5 L

a)

The no. of moles of N₂O₅ after 6 min is,

The reaction is first order, so the integrated rate of equation of first order is

ln (A/ A₀) = - kt

Where, A is the concentration, A₀ is an initial concentration and k is the constant and t is the time.

ln [(n_N2O5) / 2.3 * 10^-2 ] = - 6.82 * 10^-3* 6 * 60

ln [(n_N2O5) / 2.3 * 10^-2 ] = - 2.455

By applying exponential on both sides, we get

(n_N2O5) / 2.3 * 10^-2 = e ^{- 2.455}

(n_N2O5) / 2.3 * 10^-2 = 0.0859

n_N2O5 = 1.976 * 10^-3

b)

By using,

ln ( 1.9 * 10^-2 / 2.3 * 10^-2 ) = - 6.82 * 10^-3 * t

ln (0.826) = - 6.82 * 10^-3 * t

t = 0.1912 / 6.82 * 10^-3

t = 28.04 sec

= 0.467 min

c)

The half life of N₂O₅ at 70 ^o

For first order reaction ;

t ½ = 0.693 / k

= 0.693 / 6.82 * 10^-3

= 101.61 sec

= 1.69 min