If compound X has a first-order half-life of 24 seconds, how long would you have to wait for only 33% of the original material to be left? Explain why not knowing original Molarity of X does not effect this problem.
k = 0.693/t1/2
Substitute k into the below equation.
ln(No/N) = kt
No = 100%
N = 33%
k from above
Solve for t = time in seconds.
0.693/In(.100/.33)= -.5804
No. You never calculated k.
Also you know that a negative time makes no sense.
-0.693/
In(.100/.33) = 17.23 seconds
I know I did my math off but I'm not sure if I did it correct this time would it be 17.23 sec
Still no. You still did not calculate k.
k = 0.693/t1/2
t1/2 from the problem is 24 seconds.
k = 0.693/24 = ?
Then substitute k into the ln(No/N) = kt formula and solve for t in seconds.
K=0.693/24 = .028875
In(.100/.33)/-.028875= 41.34 sec
Almost but not quite.
You changed by ln 100/33 to ln 0.1/0.33 and that's the error. You divided by 100 so ln 100/33 or ln 1.00/0.33 would be right. Again, that ln 0.100/0.33 gives you a negative number and that gives a negative time which obviously can't be right.
ln(1.00/0.33) = 0.0288t so
t = about 38 seconds.
To determine the time required for only 33% of the original material to be left, we need to use the concept of half-life.
The half-life of a first-order reaction is the time it takes for half of the initial quantity of a reactant to decay. In this case, the compound X has a first-order half-life of 24 seconds.
To find the time required for 33% of the original material to be left, we need to consider the number of half-lives that have elapsed.
We can use the following formula to calculate the number of half-lives:
n = (log(R) - log(P)) / log(1/2)
where n is the number of half-lives, R is the remaining quantity (33% or 0.33 in decimal form), and P is the initial quantity (1 or 100% in decimal form).
Substituting the values, we have:
n = (log(0.33) - log(1)) / log(1/2)
n ≈ 0.898
Since the number of half-lives should be a whole number, we need to round it up to the nearest whole number, which gives us 1.
Thus, we can conclude that we need to wait for approximately one 24-second half-life for 33% of the original material to be left.
Now, let's address why not knowing the original molarity of X does not affect this problem. The original molarity is not required to calculate the time required for a given percentage of material to be left. The half-life is a characteristic property of the compound, independent of its initial concentration. As long as the compound behaves as a first-order reaction, knowing the original molarity is unnecessary.