Question

10.) Gaseous azomethane, CH3N=NCH3, decomposes in a first-order reaction when heated:

CH3N=NCH3(g) arrow N2(g)+C2H6(g)

The rate of the constant for this reaction at 600 K is 0.0216 min-1

A.) If the initial quantity of azomethane in the flask is 3.42g how much remains after 0.0590 hour?

B.) what mass N2 is formed in this time?

Answer 1

Answer: The reaction of azomethane decomposition

CH3N=NCH3(g) ------> N2(g)+C2H6(g)

Using the integrated rate law expression Kt

ln [A]o - ln [A]t= K t
where [A]_o is initial contration of A,
   [A]t = concentration A at time t
   t= time
   K = rate constant

ln ([A]o/[A]t] = K t

Putting the values given
ln (3.42/[A]t] = 0.0216 x 0.0590 x 60 {convert hours to minutes}

ln (3.42/[[A]t] = 0.76464
3.42/[A]t = exponential (0.76464)
3.42/[A]t= 1.0795
[A]t = 3.168

3.168g is the mass of azomethane left after 0.0590 hours.

Now, what mass of N2 is formed in this time?

Number of moles of azomethane consumed will be equal to the number of moles of N2 formed.

Mass of azomethane consumed = 3.42- 3.168 = 0.252g
moles of azomethane consumed = mass/molecular weight = 0.252/59 = 0.00427 moles

Number of moles of azomethane = number of moles of N2 formed
Mass of N2 = no. of moles x molecular weight of N2 = 0.00427 x 28 = 0.1196 g