Problem:
In a first order decomposition reaction, 50% of a compound decomposes in 10.5 minutes. What is the rate constant of the reaction?
The solution my professor gave says:
k = .693 / 10.5min
Where did the .693 come from?
Thanks
That is the natural logarithm of 2. You can do it this way.
Let's start with 100 atoms of something.
No = 100. Since it is 50% decomposed in 10.5 min, then 10.5 is the half-life and N at the end of that time will be 50.
So ln(No/N) = kt
ln(100/50) = k*t1/2
ln(2) = k*t1/2
0.693 = k*t1/2 and
k = 0.693/t1/2
The value of 0.693 comes from the natural logarithm of 2, which is approximately equal to 0.693. This value is frequently used in first-order kinetics calculations.
To understand why 0.693 is used in this scenario, we need to consider the nature of first-order decomposition reactions. In a first-order reaction, the rate of the reaction depends solely on the concentration of the reactant. The rate of a first-order reaction can be expressed as:
rate = -k[A]
Where:
- rate represents the rate of the reaction
- k is the rate constant
- [A] represents the concentration of the reactant
In the case of a first-order decomposition reaction, we can express the fraction of the reactant remaining as:
[A] / [A]0 = e^(-kt)
Where:
- [A]0 is the initial concentration of the reactant
- t is the time
- e is Euler's Number, approximately equal to 2.71828
At the point when 50% of the compound decomposes, the concentration of the reactant is halved ([A] / [A]0 = 0.5). We can substitute this value into the equation above:
0.5 = e^(-kt)
To solve for k, we need to isolate it. Taking the natural logarithm of both sides, we get:
ln(0.5) = -kt
As a result, the value of k is equal to:
k = ln(0.5) / -t
Since ln(0.5) is approximately equal to -0.693, we have:
k ≈ -0.693 / t
Here, the negative sign indicates that the concentration of the reactant decreases over time.
However, in your professor's solution, k is positive, indicating that the rate constant is expressed as a positive value. To achieve this, the negative sign is canceled out, resulting in:
k = 0.693 / t
Therefore, in the given scenario, the rate constant of the reaction is obtained by dividing 0.693 by the time of 10.5 minutes, as stated by your professor.