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?
k = 0.693/t1/2
ln(No/N) = kt
Use No = 100 (you can choose any number)
N = 100 x 0.011 = ?
k from above.
t = solve for this.
DrBob222 explain better please
So what is it? or how to guide in filling in the date in the equation?
Thank you DrBob222,
the fraction left after t seconds is (1/2)^(t/1234)
(1/2)^(t/1234) = 0.011
t = 8028.84
To determine the time it takes for the concentration of N2O5 to fall to 1.1% of its initial value, we need to use the half-life of the reaction.
First, let's understand what a half-life is. In a first-order reaction, the half-life is the time it takes for the concentration of the reactant to decrease by half.
Given that the half-life (t1/2) of the reaction is 1234 seconds, we can use the following formula to calculate the time required for any fraction of the reactant to be consumed:
t = t1/2 * (ln(initial concentration / final concentration) / ln(2))
We have the initial concentration and we need to find the time when the concentration is 1.1% (0.011) of the initial concentration.
Substituting the values into the formula, we get:
t = 1234 seconds * (ln(1 / 0.011) / ln(2))
Now let's calculate the time:
t = 1234 seconds * (ln(90.9) / ln(2))
Using a calculator, we can find the natural logarithm (ln) of 90.9 and divide it by the natural logarithm of 2. Then multiply it by 1234 seconds to get the result.
t ≈ 13082.73 seconds
Therefore, it will take approximately 13082.73 seconds for the concentration of N2O5 to fall to 1.1% of its initial value.