The rate constant for this first‑order reaction is 0.340 s−1 at 400 ∘C.
A⟶products
How long, in seconds, would it take for the concentration of A to decrease from 0.620 M to 0.280 M?
ln (No/N) = kt
No = 0.620
N = 0.280
k = 0.340
Substitute and solve for t (in seconds)
Well, this is a first-order reaction, which means that the rate of the reaction is proportional to the concentration of the reactant. So we can use the first-order rate equation to solve this problem.
ln( [A]t / [A]0 ) = -kt
Where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time it takes for the concentration to decrease.
Now we can rearrange this equation to solve for t:
t = -ln( [A]t / [A]0 ) / k
Substituting the values given in the problem, we have:
t = -ln(0.280 / 0.620) / 0.340
Calculating this, we find:
t ≈ 3.121 seconds
So it would take approximately 3.121 seconds for the concentration of A to decrease from 0.620 M to 0.280 M.
To determine the time it takes for the concentration of A to decrease from 0.620 M to 0.280 M, we can use the first-order integrated rate equation:
ln([A]t/[A]0) = -kt
where [A]t is the concentration of A at time t, [A]0 is the initial concentration of A, k is the rate constant, and t is the time.
Rearranging the equation to solve for t, we have:
t = (ln([A]t/[A]0)) / -k
Substituting the given values:
[A]0 = 0.620 M
[A]t = 0.280 M
k = 0.340 s^(-1)
t = (ln(0.280 M / 0.620 M)) / -0.340 s^(-1)
Calculating the value:
t ≈ (ln(0.45)) / -0.340 s^(-1)
Using a calculator:
t ≈ (−0.798) / -0.340 s^(-1)
t ≈ 2.350 s
Therefore, it would take approximately 2.350 seconds for the concentration of A to decrease from 0.620 M to 0.280 M.
To calculate the time it takes for the concentration of A to decrease from 0.620 M to 0.280 M, we need to use the integrated rate law for a first-order reaction:
ln[A]t/[A]0 = -kt
Where:
[A]t is the concentration of A at time t
[A]0 is the initial concentration of A
k is the rate constant
We can rearrange the equation to solve for time:
t = (ln([A]t/[A]0)) / -k
Now, let's plug in the given values:
[A]0 = 0.620 M (initial concentration)
[A]t = 0.280 M (final concentration)
k = 0.340 s^(-1) (rate constant)
t = (ln(0.280 M / 0.620 M)) / -0.340 s^(-1)
Simplifying further:
t = (ln(0.45)) / -0.340 s^(-1)
Using a calculator, take the natural logarithm (ln) of 0.45 and divide it by -0.340:
t ≈ 2.21 seconds
Therefore, it would take approximately 2.21 seconds for the concentration of A to decrease from 0.620 M to 0.280 M.