a first order reaction has a fifth life of 20 seconds ( the amount of time needed to use up 1/5)
a) Prove that the reaction has a constant 1/5 life.
b) find the half life.
Can you do this?
ln(No/N) = kt.
Just start with a convenient number like 100 for No; therefore, at the end of 20 s we should have 80 left. Calculate k.
ln(100/80) = k(20).
solve for k.
Then try the second 20 s span. That should leave you with 80-16 = 64. Plug
ln(80/64) = k(20) and see if k is the same thing.
For the half life, use
k = 0.693/t1/2 and solve for t1/2
how did you get 80 - 16?
oh i see 1/5 times 80 is 16 so that is the fifth life of the rxn. for the half life usage should i take 50 seconds using the convenient number of 100 seconds?
No, I think you misunderstood. If the fifth life (I've never heard it used that way) is 20 s and that's the tim for 1/5 of it to be used, then start with No = 100 atoms/whatever it is so at the end of 20 s we will have 80 remaining. Then 1/5 x 80 = 16 so we have 80-16 - 64 remaining. The 100 atoms/whatever is the number we are choosing for convenience. I chose 100 because that's easy to divide evenly. You could choose any number and it will work. The half live is calculated by using the k you get from those original calculations and pluging that into k = 0.693/t1/2 and solving for t1/2.
I get something like 60 seconds for the half life (but that isn't an exact number).
when i chose the next interval of a fifth i did 64 x 1/5 = 12.8 and then i took ln(64/12.8) and that did not give me the same constant as the first two so does this mean that the fifth life is not constant
You didn't do it right.
If you want to use 64, then 64/5 = 12.8 and 64-12.8 = 51.2, then
ln(64/51.2) = k(20)
and k DOES remain the same.
No is what you start with.
N is what remains.
12.8 is what was used, not what remains.
i think you made a mistake in the first answer you take ln(100/20) not 100/80 then it will prove it has a constant fifth life
If your definition of a fifth life is correct (the amount of time it takes to use up 1/5 of the material), I didn't make a mistake.
If we take 100 as No, then 1/5 of that is 20 and 100-20=80 so
ln(100/80) = k(20)
and it IS a constant. You get different numbers if you use 100 and 20.
ln(100/80) = 20k
k = 0.01116.
ln(80/64) = 20k
k = 0.01116
ln(64/51.2) = 20k
k = 0.01116
ln(35/(35-7) = 20k
ln(35/28) = 20k
k = 0.01116
nuff said?
a) To prove that the reaction has a constant 1/5 life, we need to show that regardless of the initial amount of reactant, it always takes the same amount of time to use up 1/5 of the substance.
Let's assume the initial amount of reactant is A₀ and the amount of reactant remaining after time t is A(t). According to a first-order reaction, the rate of reaction is proportional to the amount of reactant remaining:
-d[A(t)]/dt = k[A(t)]
Where k is the rate constant and [A(t)] represents the amount of reactant at time t.
We are given that the 1/5 life of the reaction is 20 seconds. This means that at t = 20 seconds, only 1/5 of the reactant remains:
[A(t=20 seconds)] = (1/5)A₀
Now, let's find the time at which only 1/5 of the reactant remains for any initial amount A₀. We can set up the following equation:
(1/5)A₀ = A₀ * e^(-kt)
Dividing both sides by A₀ and taking the natural logarithm (ln) of both sides:
ln(1/5) = -kt
Simplifying the equation:
ln(1/5) = -kt
We can see that the left side of the equation is a constant, ln(1/5), and the right side has the product of the rate constant k and time t. This implies that the 1/5 life of the reaction is constant, regardless of the initial amount of reactant.
b) To find the half-life, we need to determine the time it takes for the initial amount of reactant to decrease by half, or [A(t=0.5 * half-life)] = 0.5A₀.
Using the same first-order reaction equation:
[A(t=0.5 * half-life)] = A₀ * e^(-k(0.5 * half-life))
Substituting [A(t=0.5 * half-life)] = 0.5A₀:
0.5A₀ = A₀ * e^(-k(0.5 * half-life))
Dividing both sides by A₀:
0.5 = e^(-k(0.5 * half-life))
Taking the natural logarithm (ln) of both sides:
ln(0.5) = -k(0.5 * half-life)
Simplifying the equation:
ln(0.5) = -0.5k * half-life
Therefore, the half-life of the reaction can be found by solving for half-life:
half-life = 0.693 / k
Where 0.693 is the natural logarithm of 0.5.