This problem is 1st order for half life.
How long would it take for the concentration of cyclopropane to decrease to 50% of its initial value? to 25% of its initial value?
Rate constant is 9.2/s
I set up like this...
t1/2= 0.693/9.2/s= 0.075s at 50%
I can't figure out how to get the answer of 0.151s from 25%...
if it reduces to 1/2 in .075s, it will reduce to 1/4 in twice that, or .151s.
The half-life is 0.075s, so since 1/4 = 2(1/2), it takes 2 half-lives to fall to 25%.
To determine the time it takes for the concentration of cyclopropane to decrease to a certain percentage of its initial value, you can use the exponential decay equation. The equation for first-order decay is:
N(t) = N₀ * e^(-kt),
where N(t) is the concentration at time t, N₀ is the initial concentration, k is the rate constant, and e is Euler's number (approximately 2.71828).
In this case, you want to calculate the time it takes for the concentration to decrease to 50% and 25% of its initial value. Let's start with the calculation for 50%:
0.50N₀ = N₀ * e^(-kt).
Dividing both sides of the equation by N₀:
0.50 = e^(-kt).
Taking the natural logarithm (ln) of both sides to solve for t:
ln(0.50) = -kt.
Rearranging the equation to isolate t:
t = -ln(0.50) / k.
Now, substituting the given rate constant of 9.2/s into the equation:
t = -ln(0.50) / 9.2.
Evaluating this expression will give you the time it takes for the concentration to decrease to 50% of its initial value.
Now, let's move on to calculating the time it takes for the concentration to decrease to 25% of its initial value. We'll follow a similar approach:
0.25N₀ = N₀ * e^(-kt).
Dividing both sides by N₀ and rearranging the equation:
e^(-kt) = 0.25.
Again, taking the natural logarithm of both sides:
-ln(0.25) = -kt.
Isolating t:
t = -ln(0.25) / k.
Now, substitute the given rate constant of 9.2/s into the equation:
t = -ln(0.25) / 9.2.
Evaluating this expression will give you the time it takes for the concentration to decrease to 25% of its initial value. The value of t in this case should be 0.151s.