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

Answer:

The initial concentration of ethanal was 0.1590 mol/L.

Explanation:

Integrated rate law for second order kinetic:

$k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})$

k = Rate constant =$6.73\times 10^{-3} L mol s$

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:

$6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})$

$[A]_o=0.1590 mol/L$

The initial concentration of ethanal was 0.1590 mol/L.