Question

In the first order decomposition of acetone at 500°c, ch3och3→ productit is found that the concentration of acetone is 0.0300 m after 200 min and 0.0200 m after 400 min. calculate the rate constant, the half life, and the initial concentration.

Answer 1

For a first-order decomposition, the formula is: ln(x0 / x) = kt
At t = 200, x = 0.0300 M: ln(x0 / 0.03) = 200k
At t = 400, x = 0.0200 M: ln(x0 / 0.02) = 400k
Multiplying the first equation by 2: 2ln(x0 / 0.03) = 400k, and 400k is equivalent to the second equation, so:
2ln(x0 / 0.03) = ln(x0 / 0.02)
(x0 / 0.03)^2 = x0 / 0.02
x0 = 0.045 M (initial concentration)
Substituting into the 1st equation: ln(0.045 / 0.03) = 200k
k = 0.0020273 (rate constant)
The half life can be found when x = 0.5x0:
ln(x0 / 0.5x0) = 0.0020273t
ln(2) = 0.0020273t
t = 341.90 minutes (half-life)