The decomposition of hydrogen peroxide is first order in H2O2:
2H2O2(aq) → 2H2O(l) + O2(g) Rate = k [H2O2]
How long will it take for half of the H2O2 in a 10-gal sample to be consumed if the rate constant for this reaction is 5.6 x 10-2 s-1?
ln(No/N) = akt
ln(10/5) = 2*0.056t
Solve for t.
6.2
To determine the time it takes for half of the H2O2 in a 10-gal sample to be consumed, we can use the first-order rate equation:
ln([H2O2]t/[H2O2]0) = -kt,
where [H2O2]t is the concentration of H2O2 at time t, [H2O2]0 is the initial concentration of H2O2, k is the rate constant, and t is the time.
We need to rearrange the equation to solve for t:
ln([H2O2]t/[H2O2]0) = -kt
[H2O2]t/[H2O2]0 = e^(-kt)
t = - (1/k) * ln([H2O2]t/[H2O2]0)
Given:
Initial concentration, [H2O2]0 = 10 gals
Rate constant, k = 5.6 x 10^-2 s^-1
Let's calculate the time it takes for half of the H2O2 to be consumed.
To calculate the time it takes for half of the H2O2 in a 10-gallon sample to be consumed, we can use the first-order rate equation:
ln([H2O2]t/[H2O2]0) = -kt
Where:
[H2O2]t is the concentration of H2O2 at time t
[H2O2]0 is the initial concentration of H2O2
k is the rate constant of the reaction
t is the time
Since we want to find the time when half of the H2O2 is consumed, we can substitute [H2O2]t = [H2O2]0 / 2 into the equation. Rearranging the equation, we get:
ln(1/2) = -k * t
Now we can solve for t:
t = -ln(1/2) / k
Given that the rate constant (k) is 5.6 x 10^(-2) s^(-1), we can substitute this value into the equation and calculate the time.