Half-life
N2O5 dissolved in CCl4 decomposes to give N2O4 and O2. The reaction is first order with a half-life of 1234 seconds. How long will it take, in seconds, for the concentration of N2O5 to fall to 1.1% of its initial value?
Didn't I do this a couple of days ago for you?
k = 0.693/t1/2
Substitute and solve for k.
ln(No/N) = kt
Set any convenient number, such as 100, for No.
N, then, is 1.1% No. For No = 100, N = 1.10
k = from above.
t = solve for t in seconds.
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Thank you DrBob222,
the fraction left after t seconds is (1/2)^(t/1234)
(1/2)^(t/1234) = 0.011
t = 8028.84
To find out how long it will take for the concentration of N2O5 to fall to 1.1% of its initial value, we can use the concept of half-life.
First, we need to determine the rate constant (k) for the reaction. The reaction is first order, so the half-life (t1/2) is related to the rate constant as follows:
t1/2 = (0.693 / k)
Given that the half-life is 1234 seconds, we can rearrange the equation to solve for the rate constant:
k = 0.693 / t1/2
Plugging in the given value of the half-life:
k = 0.693 / 1234
Next, we can use the rate constant to calculate the time it takes for the concentration of N2O5 to decrease to 1.1% of its initial value. The equation for a first-order reaction over time (t) is:
[N2O5]t = [N2O5]0 * e^(-kt)
Where [N2O5]t is the concentration of N2O5 at time t, [N2O5]0 is the initial concentration, k is the rate constant, and e is the base of the natural logarithm (approximately 2.71828).
In this case, we want to solve for t when [N2O5]t is 1.1% (or 0.011) of [N2O5]0:
0.011 = [N2O5]0 * e^(-kt)
Rearranging the equation, we can solve for t:
t = (-1 / k) * ln(0.011 / [N2O5]0)
Plugging in the determined rate constant (k) and assuming the initial concentration ([N2O5]0) is 100% (1.0):
t = (-1 / 0.693 / 1234) * ln(0.011 / 1.0)
Simplifying the equation, we can calculate the time it will take:
t ≈ 8871 seconds
Therefore, it will take approximately 8871 seconds for the concentration of N2O5 to fall to 1.1% of its initial value.