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 2.90×10−2 mol of N2O5(g) in a volume of 1.7 L.
Part A How many moles of N2O5 will remain after 4.0 min ?

Answer 1

Answer:

0.00564 moles

Explanation:

Given that:

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

Initial concentration [A₀] = $2.90\times 10^{-2}$ mol

Time = 4.0 min = $4.0\times 60$ sec = 240 sec

Using integrated rate law for first order kinetics as:

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

Where,  

$[A_t]$ is the concentration at time t

So,  

$[A_t]=2.90\times 10^{-2}\times e^{-6.82\times 10^{-3}\times 240}=2.9\times \frac{1}{10^2}\times \frac{1}{e^{1.6368}}$

$[A_t]=0.00564\ moles$