Acetone is one of the most important solvents in organic chemistry, used to dissolve everything from fats and waxes to airplane glue and nail polish. At high temperature, it decomposes in a first order process to methane and ketene (CH2=C=O).At 600C, the rate constant is 8.7x10^-3-s^-1
(a) what is the half-life of the reaction?
(b) how much time is required for 40% of a acetone to decompose?
(c) how much time is required for 90% of a sample
Of a acetone to decompose?
To solve the given problem, we will use the first-order reaction kinetics equation:
ln(A/A₀) = -kt
where:
A is the concentration of acetone at time t,
A₀ is the initial concentration of acetone,
k is the rate constant,
t is the time.
(a) Half-life of the reaction:
The half-life (t₁/₂) is the time taken for the concentration of acetone to reduce to half its initial value. Since A = 0.5A₀ when t = t₁/₂, we can rearrange the equation as follows:
ln(0.5) = -kt₁/₂
Solving for t₁/₂:
t₁/₂ = -ln(0.5)/k
Substituting the given value of k = 8.7x10^-3 s^-1, we find:
t₁/₂ = -ln(0.5)/(8.7x10^-3) s
t₁/₂ ≈ 79.3 seconds (rounded to two decimal places)
Therefore, the half-life of the reaction is approximately 79.3 seconds.
(b) Time required for 40% of acetone to decompose:
To find the time required for 40% of acetone to decompose, we need to determine the concentration of acetone at that time (A):
ln(A/A₀) = -kt
Substituting the given value of A/A₀ = 0.40 (40%), we can rearrange the equation as follows:
ln(0.40) = -kt
Solving for t:
t = -ln(0.40)/k
Substituting the given value of k = 8.7x10^-3 s^-1, we find:
t ≈ -ln(0.40)/(8.7x10^-3) s
t ≈ 47.7 seconds (rounded to two decimal places)
Therefore, it would take approximately 47.7 seconds for 40% of acetone to decompose.
(c) Time required for 90% of acetone to decompose:
Follow the same steps as in part (b) to find the time required for 90% of acetone to decompose:
ln(A/A₀) = -kt
Substituting the given value of A/A₀ = 0.90 (90%), we can rearrange the equation as follows:
ln(0.90) = -kt
Solving for t:
t = -ln(0.90)/k
Substituting the given value of k = 8.7x10^-3 s^-1, we find:
t ≈ -ln(0.90)/(8.7x10^-3) s
t ≈ 183.5 seconds (rounded to two decimal places)
Therefore, it would take approximately 183.5 seconds for 90% of acetone to decompose.
To solve these questions, we need to understand the concept of the half-life of a reaction and use the given rate constant to calculate the time required for a certain percentage of acetone to decompose.
(a) The half-life of a reaction is the time it takes for half of the initial concentration of reactant to be consumed. In a first-order reaction, the half-life is calculated using the expression:
t1/2 = (0.693 / k)
where t1/2 is the half-life of the reaction, and k is the rate constant.
Given that the rate constant (k) is 8.7x10^-3 s^-1, we can calculate the half-life as follows:
t1/2 = (0.693 / 8.7x10^-3) = 79.31 seconds (approximately)
Therefore, the half-life of the reaction is approximately 79.31 seconds.
(b) To calculate the time required for 40% of the acetone to decompose, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, and k is the rate constant.
By rearranging the equation, we get:
t = -ln([A]t/[A]0) / k
The concentration of the reactant, [A]t, is given as 40% or 0.40 of the initial concentration ([A]0).
Plugging in the values, the equation becomes:
t = -ln(0.40) / (8.7x10^-3)
Calculating this expression:
t ≈ 31.89 seconds
Therefore, it would take approximately 31.89 seconds for 40% of the acetone to decompose.
(c) Similarly, to calculate the time required for 90% of a sample of acetone to decompose, we can use the same equation:
t = -ln([A]t/[A]0) / k
The concentration of the reactant, [A]t, is given as 90% or 0.90 of the initial concentration ([A]0).
Plugging in the values, the equation becomes:
t = -ln(0.90) / (8.7x10^-3)
Calculating this expression:
t ≈ 104.89 seconds
Therefore, it would take approximately 104.89 seconds for 90% of the sample of acetone to decompose.
a. k=0.693/t1/2. Substitute and solve for half life.
b. ln(No/N) = kt
No = 100
N = 60 (that's 100-40 = 60)
k = from above.
t = solve for this.
c. See b.