The half life for a first order reaction is know to be 4.5 h. How long will it take for 85% to be decomposed, in h?
k = 0.693/t1/2 and substitute into the equation below.
ln(No/N) = kt
No I would use 100
N then will be 85
k from above.
Solve for t in hours.
Well, if the half-life of a first order reaction is 4.5 hours, we can calculate how many half-lives it would take for 85% of the substance to decompose. But hey, let's not get so serious about it.
Imagine you're at a party, and people keep leaving every 4.5 hours. It's like a reverse game of musical chairs, but instead of sitting down, everyone vanishes! Anyway, after the first 4.5 hours, you still have 50% of the people remaining. After the second 4.5 hours, half of those remaining people leave, so you have 25% left.
We keep going, after the third 4.5 hours, we lose another half, so we have 12.5% left. After the fourth 4.5 hours, half of 12.5% would leave, leaving us with just 6.25%.
Now, to find when 85% of people would leave, we need to find out how many 4.5 hour intervals it would take for us to be left with just 15% (100% - 85%).
So, if we divide 100% by 6.25%, we get 16. And since each division represents a 4.5-hour interval, we simply multiply 16 by 4.5.
Therefore, it would take approximately 72 hours for 85% of the substance to decompose. See? Chemistry can be fun!
To determine the time it will take for 85% of the substance to decompose in a first-order reaction using the half-life, you can use the following formula:
t = (t1/2 / ln(2)) * ln(C₀ / Cₜ)
Where:
t = time it will take for 85% to decompose
t1/2 = half-life of the reaction
C₀ = initial concentration
Cₜ = final concentration (85% decomposed)
Given that the half-life (t1/2) is known to be 4.5 h, and we want to find the time for 85% (Cₜ) to decompose, let's substitute the values into the formula:
t = (4.5 / ln(2)) * ln(C₀ / Cₜ)
Since we're dealing with the decomposition of the substance, the concentration at the start (C₀) would be 100%, and the final concentration at 85% decomposed (Cₜ) would be 15%.
t = (4.5 / ln(2)) * ln(100% / 15%)
Now, let's solve for t using a calculator:
t ≈ 9.83 hours
So, it will take approximately 9.83 hours for 85% of the substance to be decomposed.
To find the time it takes for a certain percentage of a substance to decompose in a first-order reaction, we can use the formula:
t = (ln(initial concentration / final concentration)) / k
where:
- t is the time it takes for the reaction to occur,
- initial concentration is the starting concentration of the substance,
- final concentration is the desired concentration at which we want to find the time, and
- k is the rate constant for the first-order reaction.
In this case, we are given the half-life of the reaction, which can be used to derive the rate constant (k). The half-life (t1/2) can be defined as the time it takes for half of the initial concentration to decompose.
t1/2 = 0.693 / k
Given that the half-life (t1/2) is 4.5 hours, we can rearrange the equation to solve for the rate constant (k):
k = 0.693 / t1/2
= 0.693 / 4.5
≈ 0.154 h^-1
Now that we know the rate constant (k), we can solve for the time it takes for 85% of the substance to decompose:
t = (ln(initial concentration / final concentration)) / k
Let's assume the initial concentration is 100%, and we want to find the time at which 85% remains:
t = (ln(100% / 15%)) / 0.154
= (ln(6.6667)) / 0.154
≈ 7.97 hours
Therefore, it will take approximately 7.97 hours for 85% of the substance to decompose in this first-order reaction.