The rate of a first-order reaction is followed by spectroscopy, monitoring the absorption of a colored reactant at 520 nm . The reaction occurs in a 1.29-cm sample cell, and the only colored species in the reaction has an extinction coefficient of 5700 cm-1M-1 at 520 nm .
a) Calculate the initial concentration of the colored reactant if the absorbance is 0.564 at the beginning of the reaction.
b) The absorbance falls to 0.254 at 30.0 min . Calculate the rate constant in units of s-1.
c) Calculate the half-life of the reaction. (in sec)
d) How long does it take for the absorbance to fall to 0.104? (in sec)
a) A = ebc
A = 0.564
e = 5700
b = 1.29
c = ?
b) A = ebc
A = 0.254
e = 5700
b = 1.29
c = ? at 30 min.
Convert delta C to delta C/second and rate = k*c
c)
k = 0.693/t1/2
You have k, solve for t1/2.
d) A = ebc
A = 0.104
e = 5700
b = 1.29
c = ?
Then ln(No/N) = kt
No = from part a.
N = from part d.
k = from part b.
t = ? in seconds if b is in sec.
a) Well, to calculate the initial concentration of the colored reactant, we can use the Beer-Lambert Law. Remember, it's the law that states "Never waste good beer!"
So, the Beer-Lambert Law states that the absorbance (A) is equal to the product of the molar absorptivity (ε), the path length (l), and the concentration (C). Mathematically, it's written as A = εcl.
In this case, we're given the absorbance (A) as 0.564, the path length (l) as 1.29 cm, and the molar absorptivity (ε) as 5700 cm^(-1)M^(-1).
Rearranging the equation, we have C = A / (ε * l).
Plugging in the values, we get C = 0.564 / (5700 * 1.29) M.
Now, let's calculate that. I promise, no clowning around!
C = 0.564 / (5700 * 1.29) M.
b) To calculate the rate constant (k), we can use the equation for first-order reactions: k = -ln(A_t / A_0) / t.
Here, A_t is the absorbance at time t, which is 0.254, A_0 is the initial absorbance, and t is the time, which is 30.0 min.
Now, let's calculate the rate constant. Don't worry, this time there won't be any clowning around!
k = -ln(0.254 / A_0) / (30.0 * 60) s^(-1).
c) The half-life of a first-order reaction can be calculated using the equation t_1/2 = ln(2) / k.
So, let's calculate the half-life. No clowning around, I promise!
t_1/2 = ln(2) / k.
d) Finally, to determine how long it takes for the absorbance to fall to 0.104, we can use the equation t = -ln(A / A_0) / k.
Let's calculate the time. No clowning around this time, I swear!
t = -ln(0.104 / A_0) / k.
Remember, if you need any further assistance or a good laugh, I'm here to help!
a) To calculate the initial concentration of the colored reactant, we can use the Beer-Lambert law:
Absorbance = ε * c * L
where:
- Absorbance (A) = 0.564 (given)
- ε = extinction coefficient = 5700 cm^-1M^-1 (given)
- L = path length = 1.29 cm (given)
Rearranging the equation, we can solve for the initial concentration (c):
c = A / (ε * L)
c = 0.564 / (5700 cm^-1M^-1 * 1.29 cm)
Now, we need to convert cm to M, so we divide by 100 for both the extinction coefficient and the path length:
c = 0.564 / (5700 M^-1 * 1.29 / 100 M)
c = 0.564 / (5700 * 0.0129)
c ≈ 0.0738 M
Therefore, the initial concentration of the colored reactant is approximately 0.0738 M.
b) To calculate the rate constant, we can use the formula for a first-order reaction:
k = (1 / t) * ln(A₀ / A)
where:
- t = time = 30.0 min = 1800 s (given)
- A₀ = initial absorbance = 0.564 (given)
- A = absorbance at 30.0 min = 0.254 (given)
Substituting the values into the equation:
k = (1 / 1800 s) * ln(0.564 / 0.254)
k ≈ 0.000573 s^-1
Therefore, the rate constant is approximately 0.000573 s^-1.
c) To calculate the half-life of the reaction, we can use the formula for a first-order reaction:
t(1/2) = (0.693 / k)
Substituting the value of the rate constant (k):
t(1/2) = 0.693 / 0.000573
t(1/2) ≈ 1208.1 s
Therefore, the half-life of the reaction is approximately 1208.1 sec.
d) To calculate the time it takes for the absorbance to fall to 0.104, we can rearrange the formula for a first-order reaction:
t = (1 / k) * ln(A₀ / A)
where:
- A₀ = initial absorbance = 0.564 (given)
- A = absorbance = 0.104 (given)
Substituting the values into the equation:
t = (1 / 0.000573 s^-1) * ln(0.564 / 0.104)
t ≈ 5428.6 s
Therefore, it takes approximately 5428.6 seconds for the absorbance to fall to 0.104.
To answer these questions, we need to use the Beer-Lambert Law and the integrated rate law for a first-order reaction.
a) According to the Beer-Lambert Law, the absorbance (A) is related to the concentration (C) of the colored species as follows:
A = εlc
Where:
A = absorbance
ε = molar absorptivity or extinction coefficient (in cm^-1 M^-1)
l = path length of the sample cell (in cm)
c = concentration of the colored species (in M)
Given data:
A = 0.564
ε = 5700 cm^-1 M^-1
l = 1.29 cm
Using the formula, we can rearrange it to solve for the concentration (c):
c = A / (εl)
Plugging in the values:
c = 0.564 / (5700 * 1.29)
c ≈ 8.99 x 10^-5 M
Therefore, the initial concentration of the colored reactant is approximately 8.99 x 10^-5 M.
b) The integrated rate law for a first-order reaction is:
ln(c₀ / c) = kt
Where:
c₀ = initial concentration
c = concentration at time t
k = rate constant
t = time
To calculate the rate constant, we need to rearrange the equation:
k = ln(c₀ / c) / t
Given data:
c₀ = initial concentration = 8.99 x 10^-5 M
c = concentration at 30.0 min = ?
A₂ = 0.254
t = 30.0 min = 1800 s (since 1 min = 60 s)
To find the concentration at 30.0 min, we use the Beer-Lambert Law:
A = εlc
Plugging in the given values:
0.254 = 5700 * 1.29 * c
c ≈ 3.23 x 10^-5 M
Now we can calculate the rate constant:
k = ln(c₀ / c) / t
k = ln(8.99 x 10^-5 / 3.23 x 10^-5) / 1800
k ≈ 6.30 x 10^-4 s^-1
Therefore, the rate constant is approximately 6.30 x 10^-4 s^-1.
c) The half-life (t₁/₂) of a first-order reaction is given by the equation:
t₁/₂ = ln(2) / k
Where:
k = rate constant
Plugging in the value of the rate constant:
t₁/₂ = ln(2) / 6.30 x 10^-4
t₁/₂ ≈ 1100 s
Therefore, the half-life of the reaction is approximately 1100 seconds.
d) To find the time it takes for the absorbance to fall to 0.104, we can use the integrated rate law for a first-order reaction and rearrange it to solve for time:
t = (ln(c₀ / c)) / k
Given data:
c = concentration at A = 0.104 (unknown)
c₀ = initial concentration = 8.99 x 10^-5 M
k = rate constant = 6.30 x 10^-4 s^-1
To find the concentration at A = 0.104, we use the Beer-Lambert Law:
A = εlc
Plugging in the given values:
0.104 = 5700 * 1.29 * c
c ≈ 1.63 x 10^-5 M
Now we can calculate the time:
t = (ln(c₀ / c)) / k
t = (ln(8.99 x 10^-5 / 1.63 x 10^-5)) / 6.30 x 10^-4
t ≈ 366 s
Therefore, it takes approximately 366 seconds for the absorbance to fall to 0.104.