Half-Life from Rate Constant Calculator
Determine the half-life (t1/2) of a first-order reaction using the decay rate constant (k).
Resulting Half-Life:
Understanding Half-Life and Rate Constants
In chemical kinetics and nuclear physics, the relationship between the rate constant (k) and half-life (t1/2) is fundamental for describing first-order decay processes. Whether you are tracking radioactive decay or the metabolism of a drug in the body, these two values are inversely proportional.
How the Calculation Works
For a first-order reaction, the rate of decay is directly proportional to the concentration of the substance. The half-life is the time required for the quantity to reduce to exactly half of its initial value. Because the rate of decay is exponential, the half-life remains constant regardless of the starting amount.
- t1/2: The half-life (time).
- ln(2): The natural logarithm of 2, which is approximately 0.693147.
- k: The reaction rate constant (reciprocal time units).
Step-by-Step Example
Suppose you have a radioactive isotope with a decay constant (k) of 0.03465 per year. To find the half-life:
- Identify the rate constant: k = 0.03465 yr⁻¹.
- Divide 0.693147 by k: 0.693147 / 0.03465.
- Result: t1/2 = 20 years.
Common Units Reference
| Rate Constant Unit (k) | Resulting Half-Life Unit (t1/2) |
|---|---|
| Seconds⁻¹ (s⁻¹) | Seconds |
| Minutes⁻¹ (min⁻¹) | Minutes |
| Hours⁻¹ (hr⁻¹) | Hours |
| Years⁻¹ (yr⁻¹) | Years |
Frequently Asked Questions
Does half-life change with the initial concentration?
For first-order reactions (like radioactive decay), no. The half-life is independent of the initial concentration. This is why the formula only requires the rate constant.
What if my reaction is second-order?
This specific calculator and formula (0.693/k) apply only to first-order reactions. Second-order reactions have a half-life formula that depends on the initial concentration: t1/2 = 1 / (k[A]₀).
Why is ln(2) used?
The natural logarithm of 2 appears because the half-life is the time it takes for a substance to reach 50% (or 1/2) of its original concentration. In the integrated rate law [A] = [A]₀e⁻ᵏᵗ, when [A]/[A]₀ = 0.5, solving for t yields ln(2)/k.