Question

2 attempts lett Check my work Report problem Be sure to answer all parts. For the decomposition of gaseous dimitrogen pentaoxide, Hint 2 NOC) 4 NOC) + O2(e) Solution the rate constant is 2.8 x 10s at 60°C. The initial concentration of N,Os is 2.42 mol/L Guided Solution (a) What is [NO] after 5.00 min? mol (b) What fraction of the NO has decomposed after 5.00 min? 56 83

Answer 1

This is a first-order reaction as can be seen from the unit of rate constant which is s^-1. Now, the initial concentration of N₂O₅ given is 2.42 mol/L i.e. 2.42 M. and given k = 2.8*10^-3 s^-1.

a) We know, after some time the concentration will be decreased and that will be "a". To find out the value of a we have the expression for first-order rate law:

$kt=2.303log\frac{2.42 M}{a}$

Given t is 5 min = 300 s.

Thus, $2.8\times 10^{-3}\times 300=2.303log\frac{2.42 M}{a}$

$=>0.364=log\frac{2.42 M}{a}$

$=>\frac{2.42}{a}=2.316$

=> a = 1.0448 M.

Thus, the concentration of N₂O₅ after 5 min is = 1.0448 g/mol.

b) Thus the fraction of N₂O₅ dissociated = 1.0448/2.42 = 0.4317 = 43.17%