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📚 First Order Half Life 計算機

專業免費的 First Order Half Life 計算機。純前端本機毫秒級計算、零資料上傳、極致隱私保護。提供詳細公式推導、實例示範與常見問答。

即時推算結果 (Computed Output)
10.0021 seconds
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What Is a First-Order Half-Life?

The half-life (t½) is the time required for the concentration of a reactant to fall to half its original value. For a first-order reaction, this half-life is constant throughout the reaction and depends only on the rate constant k — not on how much reactant you start with. This makes first-order kinetics especially predictable and is why it describes radioactive decay, many drug-elimination processes, and a wide range of chemical decompositions.

How to Use This Calculator

Enter the first-order rate constant k in reciprocal seconds (1/s) and the calculator returns the half-life in seconds. If your rate constant is in other units (1/min, 1/hr), the resulting half-life will be in the matching time unit (minutes, hours), since the formula is unit-agnostic as long as you stay consistent.

The Formula Explained

Starting from the integrated first-order rate law, \ln([A]/[A]_0) = -kt, set [A] = \tfrac{1}{2}[A]_0. Solving gives:

t_{1/2} = ((\ln 2) ÷ (k)) = ((0.693) ÷ (k))

The natural log of 2 (≈ 0.693) appears because we are halving the concentration. Notice the initial concentration cancels entirely — a defining feature of first-order behavior.

Worked Example

Suppose a reaction has a rate constant k = 0.0693\ 1/s. Then

t_{1/2} = ((0.693) ÷ (0.0693)) = 10 seconds.
So every 10 seconds, the remaining reactant is cut in half: 100% → 50% → 25% → 12.5%, and so on.

使用注意事項與精準校對原則(6 大核心守則)

📐 1. 嚴格對齊輸入量綱與單位

輸入各項數值時請務必確認量綱(例如小時、天數、金額、百分比),避免單位不匹配引發推算偏差。

🔒 2. 純前端本機毫秒級隱私保障

所有計算邏輯 100% 在您的終端瀏覽器內完成,資料絕不上傳伺服器,嚴格守護商業與個人隱私。

📊 3. 符合公認學術與產業計算模型

演算法嚴格遵循國際通用數學公式與標準計算模型,經多重情境交叉核對,數值嚴密精確。

📋 4. 一鍵複製與免登入 PDF 匯出

支援一鍵複製推算結果字串至剪貼簿,亦可直接點擊列印按鈕輸出純淨 PDF 工作報表。

🔄 5. 多情景敏感度動態對比

建議針對極端值與多組預期情景分別試算,觀察輸出數據的波動區間,輔助做出更穩健的決策。

🛡️ 6. 專業諮詢與日常參考免責聲明

本工具運算結果供日常排程、學業研習與事前參考之用,如涉及重大法規或投資決策請諮詢持證專家。

常見問題 (FAQ)

Does initial concentration affect first-order half-life?
No. Unlike zero- or second-order reactions, a first-order half-life is independent of starting concentration.
Where does 0.693 come from?
It is \( \ln 2 \), the natural logarithm of 2, which arises when solving the rate law for a 50% drop.
Can I use this for radioactive decay?
Yes — radioactive decay is first order, with the decay constant λ playing the role of k, so \( t_{1/2} = \frac{0.693}{\lambda} \).