Second-order half-life calculator
A second-order rate constant of 0.5 L/(mol·s) with an initial concentration of 0.1 mol/L gives a half-life of 20 seconds, because the half-life is one over k times the starting concentration. Enter both values to see every step. Free in your browser.
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Worked example
- Second-order rate constant k
- 0.5
- Initial reactant concentration [A]₀
- 0.1
- K
- 0.5
- A0
- 0.1
- Half life
- 20
- T half
- 20
A second-order half-life calculator turns a second-order rate constant and an initial reactant concentration into the time required for the concentration to fall to half of its starting value. Enter k and [A]₀ for a process whose rate law is rate equals k times concentration squared; the tool returns t½ equals one over k times [A]₀, echoes the rounded inputs, a short formula string, step-by-step substitution, and a clear English message. Homework checks, lab kinetics reports, exam prep, and coding kata expected values all need that same closed form without opening a computer-algebra system.
What second-order half-life means for rate = k[A]²
In chemical kinetics a second-order elementary process with rate equal to k times the square of reactant concentration obeys an integrated rate law that is not exponential. The concentration of A falls according to one over concentration at time t equals one over initial concentration plus k t. Setting the concentration at time t equal to half of the initial concentration and solving for t yields the half-life formula t one-half equals one divided by k times [A] zero.
That single real number is the time at which the reactant has been consumed to fifty percent of its starting value under second-order kinetics. Unlike first-order reactions, where half-life is independent of concentration and equals the natural log of two over k, second-order half-life shortens when you start with a higher concentration and lengthens when the mixture is more dilute.
Teachers and automated graders care about the zero-input rejection as much as the happy path: a student who plugs k equals zero or [A] zero equals zero into the formula has left the domain of this tool, and a silent return of infinity would hide that mistake. Units must stay consistent: if k is reported in liters per mole per second and concentration in moles per liter, half-life comes out in seconds.
The check path recomputes the product and reciprocal with the same fixed rounding used for published fields, so a successful response always ships with steps you can assert in tests or worksheets.
Practical tips, limits, and related searches
Keep k and [A]₀ inside the published absolute range so their product and reciprocal cannot overflow into a non-finite half-life that the solver must reject. Prefer sending numbers rather than pre-formatted strings when you control the client; string numbers are accepted only when they match a strict finite numeric pattern. Near-zero inputs after floating noise are rejected so a tiny accidental zero does not produce an enormous half-life.
Prefer this capability when you already have k and [A]₀ and only need t½; if your source is a concentration-versus-time table, extract or fit k first.
What you can do with it
Homework checks for second-order kinetics
Students enter k and [A]₀ from textbook problems and compare hand-written t½ and unit consistency against a deterministic result before submitting.
Lab report cross-checks
After fitting a second-order rate constant from absorbance or concentration data, compute t½ at the experimental initial concentration for the discussion section.
FAQ
What is the second-order half-life formula?
For rate = k[A]², the half-life is t½ = 1 / (k · [A]₀). It depends on both the rate constant and the initial concentration.
How is this different from first-order half-life?
First-order t½ = ln(2)/k does not depend on concentration. Second-order t½ shortens as [A]₀ increases.
What units should I use?
Use consistent units: if k is in L/(mol·s) and [A]₀ in mol/L, t½ is in seconds. The tool does not convert units for you.
Is the second-order half-life calculator free?
The browser tool is free and runs locally.
What happens to the half-life if I double the starting concentration?
It is cut in half, because the half-life is one divided by k times the initial concentration. A more dilute mixture takes longer to lose half of its reactant.
Related tools
For developers — API access
Everything on this page is available programmatically. This section is for teams who want to wire it into their own systems; everyone else can just use the tool above.
API endpoint
Prefer to automate it? One authenticated POST creates the task; the result comes back by webhook or a signed link. The same capability also runs here on the web, by email and from Telegram — and soon from our app too.
Call it from your stack
curl -X POST https://api.kit.forhosting.com/chem/second-order-half-life \
-H "Authorization: Bearer $KIT_KEY" \
-H "Content-Type: application/json" \
-d '{"rate_constant":0.5,"initial_concentration":0.1}'const res = await fetch("https://api.kit.forhosting.com/chem/second-order-half-life", {
method: "POST",
headers: {
"Authorization": `Bearer ${process.env.KIT_KEY}`,
"Content-Type": "application/json"
},
body: JSON.stringify({
"rate_constant": 0.5,
"initial_concentration": 0.1
})
});
const { task_id } = await res.json();import os, requests
res = requests.post(
"https://api.kit.forhosting.com/chem/second-order-half-life",
headers={"Authorization": f"Bearer {os.environ['KIT_KEY']}"},
json={
"rate_constant": 0.5,
"initial_concentration": 0.1
},
)
task_id = res.json()["task_id"]<?php
$res = file_get_contents("https://api.kit.forhosting.com/chem/second-order-half-life", false, stream_context_create([
"http" => [
"method" => "POST",
"header" => "Authorization: Bearer " . getenv("KIT_KEY") . "\r\nContent-Type: application/json",
"content" => '{"rate_constant":0.5,"initial_concentration":0.1}',
],
]));
$task = json_decode($res, true);body := bytes.NewBufferString(`{"rate_constant":0.5,"initial_concentration":0.1}`)
req, _ := http.NewRequest("POST", "https://api.kit.forhosting.com/chem/second-order-half-life", body)
req.Header.Set("Authorization", "Bearer "+os.Getenv("KIT_KEY"))
req.Header.Set("Content-Type", "application/json")
res, _ := http.DefaultClient.Do(req)Example request
{
"rate_constant": 0.5,
"initial_concentration": 0.1
}Example response
{
"task_id": "tsk_a1b2c3d4e5f6a1b2c3d4e5f6",
"type": "chem.second_order_half_life",
"status": "queued",
"_links": {
"result": "/tasks/tsk_…/result"
}
}The API is asynchronous: the call returns a task_id immediately and the result arrives by webhook. Polling is capped at 1 req/s per task.
Pricing
Published price — no tokens, no invented credits. A failed task is never charged.
Limits
max_abs | 1000000000000 |
Errors
| HTTP | Code | Meaning |
|---|---|---|
401 | unauthorized | Missing or invalid API key. |
402 | insufficient_balance | Your balance doesn't cover the task price. |
404 | unknown_type | That task type doesn't exist. |
429 | rate_limited | Too many requests. Use the webhook instead of polling. |
Technical notes
When either the rate constant or the initial concentration is zero the half-life is undefined, so the API returns a clear invalid_input error instead of infinity or a silent NaN.
How to call the API and what comes back
Both the rate constant and the initial concentration must be nonzero; values whose absolute value is at or below one times ten to the minus fifteen are treated as zero and rejected. Send a JSON body with two required finite numbers: rate_constant (aliases k, k_constant, second_order_k, and rate_k) and initial_concentration (aliases a0, A0, concentration, c0, initial_conc, and conc0). Absolute values must stay within the published limit of one trillion so intermediate products remain finite in IEEE double precision on the Worker and in the browser. On success the response echoes rounded rate_constant and k, initial_concentration and a0, sets half_life and t_half to the same rounded t½ value, formula to the string t½ equals one over k times [A]₀, message to a short English summary, and steps to an ordered array of substitution lines that show the product k times [A]₀ and the final reciprocal. Failed validation—missing fields, non-finite numbers, values outside absolute limits, zero or near-zero k or concentration, or a non-finite half-life—returns invalid_input and is never charged. Pricing is a flat $0.002 per successful request with unit item and zero marginal unit price.
Practical tips, limits, and related searches
When teaching, pair examples: k equals zero point five with [A]₀ equals zero point one yields t½ equals twenty; k equals two with [A]₀ equals zero point five yields t½ equals one; and either field set deliberately to zero demonstrates the hard invalid_input path for division by zero. This landing targets people searching for a second-order half-life calculator, t half equals one over k A zero, or second-order kinetics half life online; it is not a first-order half-life tool, not a zero-order calculator, and not a nonlinear fit of k from a time series. Failed tasks that return invalid_input do not consume prepaid balance, which keeps exploratory scripts cheap when learners deliberately set concentration to zero. Round-trip checks are easy: recompute one over k times [A]₀ by hand and confirm it equals the published half_life within the same twelve-decimal rounding, and confirm that doubling [A]₀ halves t½ when k is held fixed.
Integration use cases
Unit tests for kinetics solvers
Generate known-good half-life cases plus deliberate zero-k and zero-concentration rejections for coding kata expected values.
Technical questions
What happens when k or concentration is zero?
The half-life is undefined (division by zero). This tool returns invalid_input instead of infinity or NaN.