In the first order decomposition of acetone at 500°c, ch3och3→ productit is found that the concentration of acetone is 0.0300 m

Question

Lena

9 days ago

10

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.

Chemistry

1 answer:

lorasvet [2.5K] · Answer 1 · 2026-03-27T12:03:55+03:00

For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).

In the first order decomposition of acetone at 500°c, ch3och3→ productit is found that the concentration of acetone is 0.0300 m

Response:

To elaborate:

Step 1: Write the balanced reaction equation

Step 2: Calculate the moles of NaOH