The rate constant for the reaction
2A → B
is 7.63 × 10−3 s−1 at 110°C. The reaction is first order in A. How long (in seconds) will it take for [A] to decrease from 1.17 M to 0.590 M?
To solve this problem, we can use the first-order rate equation:
ln[A]t = -kt + ln[A]0
Where:
[A]t = concentration of A at time t
k = rate constant
[A]0 = initial concentration of A
We are given:
k = 7.63 × 10^(-3) s^(-1)
[A]0 = 1.17 M
[A]t = 0.590 M
We can rearrange the equation to solve for time (t):
t = (ln[A]t - ln[A]0) / -k
Now let's substitute the values into the equation and calculate the time:
t = (ln(0.590) - ln(1.17)) / -(7.63 × 10^(-3))
Using a scientific calculator or computer program, we can solve this:
t = (-0.5276 - 0.1520) / -(7.63 × 10^(-3))
t = -0.6796 / -(7.63 × 10^(-3))
t = 0.6796 / (7.63 × 10^(-3))
t = 88.94 seconds
Therefore, it will take approximately 88.94 seconds for [A] to decrease from 1.17 M to 0.590 M.