In a first order decomposition reaction, 50% of a compound decomposes in 10.5 min. What is the rate constant of the reaction? How long does it take for 75% of the compound to decompose?
My professor did this problem in class:
k = .693 / 10.5 mins
= .066 min ^-1
where did .693 come from?
Thanks
it's from the half-life formula for first order process rate t1/2 = .693 / k
0.693 is the natural log of 2.
It comes from using the expression
ln(No/N) = kt
ln(100/50) = k*t1/2
ln 2 = k*t<sub<1/2
k = ln2/t1/2 and
k = 9.693/t1/2
The No/N I used above as 100 and 50 can be any number choose as long as N is 1/2 of No.
The value 0.693 comes from the natural logarithmic constant, ln(2).
In a first-order decomposition reaction, the rate of reaction follows a first-order kinetics equation, which can be expressed as:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of the compound at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time.
If 50% of the compound decomposes in 10.5 minutes, it means that [A]t/[A]0 = 0.5. Plugging these values into the equation above, we can solve for k:
ln(0.5) = -k * 10.5 min
We can rewrite this equation as:
ln(0.5) / 10.5 min = -k
To simplify the equation further, we use the fact that ln(0.5) is equal to -ln(2), so we get:
-ln(2) / 10.5 min = -k
Therefore, k = ln(2) / 10.5 min.
The natural logarithmic constant ln(2) is approximately equal to 0.693, so we substitute this value into the equation:
k ≈ 0.693 / 10.5 min
k ≈ 0.066 min^-1
Hence, the rate constant of the reaction is approximately 0.066 min^-1.
To determine how long it takes for 75% of the compound to decompose, we can rearrange the first-order kinetics equation:
ln([A]t/[A]0) = -kt
Substituting [A]t/[A]0 = 0.75 (since 75% of the compound decomposes) and rearranging for t:
t = -ln(0.75) / k
Plugging in the value of k (0.066 min^-1), we can calculate the time it takes for 75% of the compound to decompose.
The value of 0.693 that your professor used in the calculation is actually the natural logarithm of 2, denoted as ln(2). The natural logarithm of a number can be calculated using a scientific calculator or online tool.
In the context of first-order decomposition reactions, the rate constant (k) can be determined using the formula:
k = ln(2) / t(1/2)
Here, t(1/2) represents the half-life of the reaction, which is the time it takes for 50% of the compound to decompose.
In the given problem, it is stated that 50% of the compound decomposes in 10.5 minutes. Since this is the half-life of the reaction, we can substitute this value into the equation:
k = ln(2) / 10.5 min
Simplifying this expression gives:
k ≈ 0.693 / 10.5 min
So, the correct rate constant for the reaction is approximately 0.066 min^(-1).
To determine how long it takes for 75% of the compound to decompose, you can use the same formula with the updated percentage:
t = ln(3) / k
In this case, the natural logarithm of 3, denoted as ln(3), is used because we want to find the time it takes for 75% (or three halves) of the compound to decompose.
Substituting the rate constant value we calculated earlier, we get:
t ≈ 1.099 / 0.066 min^(-1)
Simplifying this expression gives:
t ≈ 16.65 minutes
Therefore, it would take approximately 16.65 minutes for 75% of the compound to decompose.