Question

The rate constant k of the second-order reaction CH3CHO→CH4+CO is 6.73×10−3Lmol s. The concentration of CH3CHO at t=50.0 seconds is 0.151 mol/L. What was the initial concentration of CH3CHO?

Answer 1

The initial concentration of ethanal was 0.1590 mol/L.

Step-by-step explanation:

Integrated rate law for second order kinetic:

`k=(1)/(t)((1)/([A])-(1)/([A]_o))`

k = Rate constant =
`6.73* 10^(-3) L mol s`

t = Time elapsed = 50.0 s

`[A]_o` =initial concentration of ethanal

[A] = Concentration of ethanal left after time t = 0.151 mol/L

On substituting the value:

`6.73* 10^(-3) L mol s=(1)/(50.0 s)((1)/(0.151 mol/L)-(1)/([A_o]))`

`[A]_o=0.1590 mol/L`

The initial concentration of ethanal was 0.1590 mol/L.