Question

The gas-phase decomposition of CH3CHO (g) occurs according to the equation

CH3CHO (g) ----> CH4 (g) +CO (g) and is second order. The value of the rate constant is 0.105 M-1 x s-1 at 490 degrees Celcius. If the concentration of CH3CHO (g) is 0.012 M initially, what will be its concentration 5.0 minutes later.

Answer 1

Concentration of CH3CHO 5.0 minutes later is

`[A] =8.71* 10^(-3)`M or 0.00871 M

Step-by-step explanation:

The concentration for second order reaction can be calculated using the formula:

`(1)/([A])=(1)/([A_(0)])+kt`

Here , [A] = Concentration of substance at time t

t = time

`[A_(0)]` = Initial concentration

k = rate constant

According to question :

[A] = Concentration of CH3CHO at time 5.0s = ?

`[A_(0)]` = 0.012 M

t = 5.0 min

Convert it into second (because rate constant units are in second)

1 min = 60 s

t = 5 x 60 = 300 s

k = 0.105 M-1 s-1

Insert these value into,

`(1)/([A])=(1)/([A_(0)])+kt`

`(1)/([A])=(1)/(0.012)+ 0.105* 300`

`(1)/([A])=83.33+ 31.5`

`(1)/([A])=83.33+ 31.5`

`(1)/([A])=114.83`

`[A] =(1)/(114.83)`

`[A] =8.71* 10^(-3)`M

Hence ,concentration of CH3CHO 5.0 minutes later is

`[A] =8.71* 10^(-3)`M or 0.00871 M