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?

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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:

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:

Where,  

is the concentration at time t  

is the initial concentration  = mol

k is the rate constant = s⁻¹

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

Thus,

0.01034  moles of reactant are left after 5 minutes.

Given that:

k = s⁻¹

The expression for half life is:-

101.63 sec is the half-life.

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