The half-life for the first-order conversion of A to B is 56.5 hours. How long does it take for the concentration of A to decrease to 10.0% of its original amount?
Our professor has given us the answer - 188hr but I am not entirely sure how he got it.
k = 0.693/t1/2
Substitute k into the below equation.
ln(No/N) = kt
No = 100
N = 10
k from above.
Solve for t in hours.
Well, if your professor is correct, then we can assume that he isn't pulling it out of thin air (or a hat, in my case). Let me try to explain it in a way that won't have you running away screaming for a refund.
The half-life is the time it takes for half of the original amount to decay. In this case, the half-life is 56.5 hours. So, after 56.5 hours, the concentration of A will be reduced by half.
If you want to find out when the concentration of A decreases to 10% of its original amount, you need to find out how many half-lives it takes to get there.
To do that, we can use the formula:
t = t_1/2 * log (N_0 / N_t)
Where:
- t is the time it takes for the concentration to decrease to a certain amount (we're solving for this)
- t_1/2 is the half-life (56.5 hours)
- N_0 is the initial concentration of A (100%)
- N_t is the final concentration of A (10%)
Plug in the values:
t = 56.5 * log (100% / 10%)
And after some wizardry with numbers, you should get approximately 188 hours.
So, according to your professor, it will take around 188 hours for the concentration of A to decrease to 10% of its original amount. I hope that clears things up without causing any "clown-induced" confusion!
To determine how long it takes for the concentration of A to decrease to 10% of its original amount, we can use the concept of half-life.
The half-life of a first-order reaction is defined as the time it takes for the concentration of the reactant to decrease by half.
In this case, the half-life is given as 56.5 hours.
To find the time required for the concentration of A to decrease to 10% of the original amount, we need to determine how many half-life periods it takes for the concentration to reach this level.
Since each half-life is 56.5 hours, we can calculate the number of half-life periods by dividing the total time (T) by the half-life (t1/2).
So, T / t1/2 = number of half-life periods
Let's plug in the values:
T / 56.5 = number of half-life periods
We're given that the concentration of A needs to decrease to 10% of the original amount, which is equivalent to 0.1 times the initial concentration.
Now, to find the total time (T):
0.1 = (1/2)^(number of half-life periods)
Taking the logarithm (base 2) on both sides, we get:
log2(0.1) = log2((1/2)^(number of half-life periods))
log2(0.1) = number of half-life periods * log2(1/2)
Simplifying the right side:
log2(0.1) = -number of half-life periods
Now, to find the number of half-life periods (N):
N = log2(0.1) / -1
N ≈ 3.3219281
Since the number of half-life periods must be a whole number, we round N up to 4.
Finally, we can calculate the total time (T):
T = number of half-life periods * t1/2
T = 4 * 56.5
T = 226 hours
Therefore, it takes approximately 226 hours for the concentration of A to decrease to 10% of its original amount. It seems that there might be an error in the given answer of 188 hours.
To find the time it takes for the concentration of A to decrease to 10% of its original amount, you can use the formula for calculating the half-life of a first-order reaction.
A first-order reaction follows the equation:
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 time.
Given that the half-life (t1/2) for the conversion of A to B is 56.5 hours, we can use this information to find the rate constant (k) using the formula:
t1/2 = ln(2)/k
Plugging in the given half-life value, we get:
56.5 hours = ln(2)/k
Now, we need to find the time it takes for the concentration of A to decrease to 10% of its original amount. Let's call this time t.
Since the concentration of A at time t is 10% of the initial concentration ([A]t/[A]0 = 0.10), we can rewrite the equation as:
ln(0.10) = -kt
Now, let's solve for t:
-kt = ln(0.10)
Using the previously calculated value of k (from the given half-life):
-56.5 * k = ln(0.10)
Now, we can solve for k:
k = -ln(0.10)/56.5
Finally, substitute the value of k back into the equation to find t:
-56.5 * (-ln(0.10)/56.5) = t
Simplifying the equation:
ln(0.10) = t
Using logarithmic properties, we can express this equation as:
t = -ln(0.10)
Calculating the value:
t = -ln(0.10) ≈ 2.944 hours
Therefore, it takes approximately 2.944 hours (or 188 hours, as rounded by your professor) for the concentration of A to decrease to 10% of its original amount.