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?
To solve this problem, we need to use the concept of half-life and the first-order reaction equation.
First, let's understand what first-order reaction means. In a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. Mathematically, we can express it as:
rate = k[A]
Where:
rate = rate of reaction
k = rate constant
[A] = concentration of reactant A
Now, the half-life of a reaction is the time it takes for the concentration of the reactant to reduce to half of its initial value. Mathematically, we can express it as:
t1/2 = (ln 2) / k
Where:
t1/2 = half-life
ln = natural logarithm (base e)
k = rate constant
Given that the half-life of the reaction is 1234 seconds, we can use this information to find the value of the rate constant (k).
t1/2 = (ln 2) / k
1234 = (ln 2) / k
Solving for k:
k = (ln 2) / 1234
Now, we need to determine how long it will take for the concentration of N2O5 to fall to 1.1% (0.011) of its initial value.
Since it's a first-order reaction, we can use the following equation to relate the concentration of N2O5 at different times:
[A]t = [A]0 * e^(-kt)
Where:
[A]t = concentration of N2O5 at time t
[A]0 = initial concentration of N2O5
e = Euler's Number (approximately 2.71828)
k = rate constant
t = time
We know that [A]t / [A]0 = 0.011, so we can rearrange the equation as:
0.011 = e^(-kt)
Taking the natural logarithm of both sides:
ln (0.011) = -kt
Now, we can solve for t:
t = (ln (0.011)) / -k
Substituting the value of k we found earlier:
t = (ln (0.011)) / -((ln 2) / 1234)
Calculating this equation will give us the time it takes for the concentration of N2O5 to fall to 1.1% of its initial value in seconds.