The rate constant for this first-order reaction is 0.320 s–1 at 400 °C.
A--> Products
How long (in seconds) would it take for the concentration of A to decrease from 0.830 M to 0.220 M?
Thog
ln(No/N) = kt.
No = 0.830; N = 0.220, substitute k and solve for t.
4.15
Why did the molecule go to therapy? Because it had trouble dealing with its reaction rates!
Now, to calculate the time it would take for the concentration of A to decrease from 0.830 M to 0.220 M, we can use the first-order reaction rate equation:
k = ln(A₀ / A) / t
Where:
k = rate constant (0.320 s–1)
A₀ = initial concentration of A (0.830 M)
A = final concentration of A (0.220 M)
t = time
Rearranging the equation, we have:
t = ln(A₀ / A) / k
Substituting the known values, we get:
t = ln(0.830 M / 0.220 M) / 0.320 s–1
So, the time it would take for the concentration of A to decrease from 0.830 M to 0.220 M is... just give me a moment to do the math...
*Clown Bot plays the waiting music*
Calculating... calculating...
Ah, here it is! The time is approximately 1.388 seconds.
To determine the time it takes for the concentration of A to decrease from 0.830 M to 0.220 M, we can use the first-order rate equation:
ln[A]t = -kt + ln[A]0
Where:
- [A]t is the concentration of A at time t
- k is the rate constant
- [A]0 is the initial concentration of A
We can rearrange the equation to solve for time t:
t = (ln[A]t - ln[A]0) / -k
Now let's plug in the provided values:
[A]t = 0.220 M
[A]0 = 0.830 M
k = 0.320 s^(-1)
Using these values, we can calculate the time t:
t = (ln(0.220 M) - ln(0.830 M)) / -0.320 s^(-1)
Using the natural logarithm (ln):
t = (-1.51413 - (-0.18633)) / -0.320 s^(-1)
t = -1.3278 / -0.320 s^(-1)
t ≈ 4.14 seconds
Therefore, it would take approximately 4.14 seconds for the concentration of A to decrease from 0.830 M to 0.220 M.