Consider the first order reaction A --> products where 25% of A disappears in 24 seconds. What is the half life of this reaction?
...i thought it would just be doubled but it isn't. i'd appreciate an explanation.
Probably I work these the long way but I do that because then I know I will get it right.
ln(No/N) = kt
I assume a value of 100 for No, and if 25% disappears, that means 75% is left so starting with 100 gives me 75 at the end of 24 seconds.
ln(100/75) = 24*k
solve for k and I get 0.01199 (I know that's too many places but I just leave those numbers in the calculator. Then for the half life
ln(100/50) = 0.01199t1/2
solve for t1/2
Check my math but I obtained 57.8 seconds which would be rounded to 58 sec if the 24 value has just 2 s.f.
Ah, the mysterious world of chemical reactions! Well, let me explain it to you in my own clownish way.
In a first order reaction, the rate of reaction is directly proportional to the concentration of the reactant. Now, the half-life is the time it takes for half of the reactant to disappear. But here's the twist: it's not a fixed time for all reactions!
In this case, if 25% of A disappears in 24 seconds, that means 75% of A is still left. So, in the next 24 seconds, another 25% will disappear, leaving us with 50% of A. And in yet another 24 seconds, another 25% will disappear, leaving us with 25% of A, and so on.
Now, to calculate the half-life, we need to find the time it takes for 50% of A to disappear. Since it takes 24 seconds to go from 75% to 50%, and another 24 seconds to go from 50% to 25%, the total time it takes for 50% to disappear is 48 seconds.
So, my friend, the half life of this reaction would be 48 seconds, not just double the time it took for 25% to disappear. Chemistry can be quite sneaky, but don't worry, I'm here to make it a bit more fun for you!
The half-life of a first order reaction is the time it takes for half of the reactant to be consumed. In this case, you know that 25% of A disappears in 24 seconds.
To find the half-life, you need to determine the fraction of A that has disappeared after the half-life. Since 25% of A has disappeared in 24 seconds, this means that 75% of A (100% - 25%) remains after the 24 seconds.
Now, let's assume t is the half-life of the reaction. After t seconds, 50% of A should disappear. So, the fraction of A that remains after t seconds is 50%.
We can set up the following equation:
75% * A_0 = 50% * A_0 * e^(-kt)
where A_0 is the initial concentration of A, k is the rate constant, and e is the base of the natural logarithm.
Simplifying the equation, we divide both sides by A_0:
0.75 = 0.5 * e^(-kt)
Now, we can solve for k by taking the natural logarithm of both sides:
ln(0.75) = -kt
Finally, we can solve for the half-life (t) by rearranging the equation:
t = -ln(0.75) / k
Calculating this value will give you the half-life of the reaction.
To determine the half-life of a first-order reaction, we can use the following formula:
t1/2 = (0.693 / k)
Where t1/2 is the half-life and k is the rate constant of the reaction.
Now, in this case, if 25% of A disappears in 24 seconds, it means we have (100% - 25%) = 75% of A remaining after 24 seconds.
To find the rate constant (k), we can use the following equation:
ln(C0 / Ct) = kt
Where C0 is the initial concentration of A and Ct is the remaining concentration of A after a given time (t).
Since 25% of A disappears, we can say that 75% of A remains after 24 seconds. So, Ct = 0.75C0.
The equation now becomes:
ln(C0 / 0.75C0) = k * 24
Simplifying, we have:
ln(1 / 0.75) = k * 24
ln(4/3) = k * 24
Now, solve for k:
k = ln(4/3) / 24
k ≈ 0.01149 s^(-1)
Finally, substitute this value of k into the formula for half-life:
t1/2 = 0.693 / k
t1/2 = 0.693 / 0.01149
t1/2 ≈ 60.36 seconds
Therefore, the half-life of the reaction is approximately 60.36 seconds.