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. Starting with 8.00×10−2 mol of N2O5(g) in a volume of 2.9 L, how many moles of reactant are left after 5 minutes? What is its half-life?

Answer 1

Answer:

0.01034  moles of reactant are left after 5 minutes.

101.63 sec is the half-life.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t  

$[A_0]$ is the initial concentration  = $8.00\times 10^{-2}$ mol

k is the rate constant = $6.82\times 10^{-3}$ s⁻¹

Time = 5 minutes = 5*60 seconds = 300 seconds ( 1 min = 60 sec)

Thus,

$[A_t]=8.00\times 10^{-2}e^{-6.82\times 10^{-3}\times 300}\ mol=0.01034\ mol$

0.01034  moles of reactant are left after 5 minutes.

Given that:

k = $6.82\times 10^{-3}$ s⁻¹

The expression for half life is:-

$t_{1/2}=\frac{\ln2}{k}$

Thus,

$t_{1/2}=\frac{\ln2}{6.82\times 10^{-3}}\ s=101.63\ s$

101.63 sec is the half-life.