For the fist order reaction A <-> B, calculate the t1/2(half life) and k(rate constant) and also fill in the blanks.
A-Concetration t-time
20 0
10 ?
5 40 sec
3 ?
? 500 sec
t1/2=? k=?
is this for ashworth??
what do you mean?
This is a first order reaction and will be answered by using Damon's answer to the second order question.
ok i got that now thank you. how will i find the answer for the blanks in the time? is there a formula for it?
To calculate the t1/2 (half-life) and k (rate constant) for a first-order reaction, we can use the integrated rate equation 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, and t is the time.
Let's start by filling in the given concentrations and times:
A-Concentration (M) t (sec)
20 0
10 ?
5 40
3 ?
? 500
To find the time at which the concentration halves, we can use the half-life equation for a first-order reaction:
t1/2 = ln(2)/k
where ln is the natural logarithm.
To calculate the rate constant (k), we can rearrange the integrated rate equation:
k = -ln([A]t/[A]0) / t
Now, let's calculate the missing values:
1. To find the time at which the concentration is halved from 20 to 10 M, we can use the half-life equation:
t1/2 = ln(2)/k
Substituting the given values:
t1/2 = ln(2)/k = ln(2)/0.693 = 0.693/0.693 = 1 sec
So, the time at which the concentration is halved is 1 second.
2. To calculate the rate constant (k) for the given values, we can use the rearranged integrated rate equation:
k = -ln([A]t/[A]0) / t
Using the concentration of A at t=0 (20 M) and t=40 sec (5 M):
k = -ln(5/20) / 40 = -ln(0.25)/40 = -(-1.386)/40 = 0.0347 s^-1
So, the rate constant (k) is 0.0347 s^-1.
Now let's fill in the remaining blanks:
A-Concentration (M) t (sec)
20 0
10 1
5 40
3 92.47 (approximately)
2.5 500
So, the remaining concentrations and times are filled in accordingly.