Earlier point on the linear elimination phase (Hours/Min)
Concentration at t1 (mg/L, µg/mL, etc.)
Later point on the linear elimination phase
Concentration at t2
Elimination Rate Constant (ke): units⁻¹
Elimination Half-Life (t1/2): units
Slope of ln(C) vs t graph:
function calculateEliminationRate() {
var t1 = parseFloat(document.getElementById('time1').value);
var c1 = parseFloat(document.getElementById('conc1').value);
var t2 = parseFloat(document.getElementById('time2').value);
var c2 = parseFloat(document.getElementById('conc2').value);
var resultBox = document.getElementById('result-display');
var errorBox = document.getElementById('error-display');
// Reset displays
resultBox.style.display = 'none';
errorBox.style.display = 'none';
// Validation
if (isNaN(t1) || isNaN(c1) || isNaN(t2) || isNaN(c2)) {
errorBox.innerText = "Please fill in all fields with valid numbers.";
errorBox.style.display = 'block';
return;
}
if (c1 <= 0 || c2 t1 && c2 >= c1) {
errorBox.innerText = "For elimination phase, concentration at the later time (C2) should be lower than C1.";
errorBox.style.display = 'block';
// We continue calculation but warn user, or return? Usually better to return to prevent confusion.
return;
}
// Calculation: ke = (ln(C1) – ln(C2)) / (t2 – t1)
// Note: Using Natural Log (Math.log)
var timeDiff = t2 – t1;
// Handle if user put t2 before t1
if (timeDiff < 0) {
timeDiff = Math.abs(timeDiff);
// Swap concentrations for calculation logic
var tempC = c1;
c1 = c2;
c2 = tempC;
}
var naturalLogDiff = Math.log(c1) – Math.log(c2);
var ke = naturalLogDiff / timeDiff;
// Half life calculation: t1/2 = 0.693 / ke
var halfLife = 0.693 / ke;
// Slope calculation (slope = -ke)
var slope = -ke;
// Display Results
document.getElementById('ke-result').innerText = ke.toFixed(4);
document.getElementById('halflife-result').innerText = halfLife.toFixed(2);
document.getElementById('slope-result').innerText = slope.toFixed(4);
resultBox.style.display = 'block';
}
How to Calculate Elimination Rate Constant from a Graph
The elimination rate constant (ke) is a fundamental parameter in pharmacokinetics that describes the rate at which a drug is removed from the body. Determining this value from a concentration-time graph is a critical skill for understanding drug behavior, dosing intervals, and clearance.
Understanding the Semi-Logarithmic Plot
Most drugs follow first-order kinetics, meaning a constant fraction of the drug is eliminated per unit of time. When you plot Plasma Concentration vs. Time on a standard Cartesian graph, the curve is exponential. However, to calculate the elimination rate constant easily, scientists convert this data onto a semi-logarithmic graph (log of concentration vs. time).
On a semi-log plot, the elimination phase appears as a straight line. The slope of this line allows us to calculate ke.
Step-by-Step Calculation Method
To calculate the elimination rate constant manually from a graph, follow these steps:
Identify the Elimination Phase: Look for the straight-line portion of the semi-log graph. This occurs after the absorption and distribution phases are complete.
Select Two Points: Pick two distinct points on this straight line. Let's call them (t1, C1) and (t2, C2).
t represents time (e.g., hours).
C represents drug concentration.
Apply the Formula: Since the relationship is linear on a semi-log scale, the slope represents the rate of change. The formula for the elimination rate constant is:
ke = (ln(C1) – ln(C2)) / (t2 – t1)
Note: "ln" stands for the natural logarithm. If you are using log base 10, you must multiply the result by 2.303.
Calculating Half-Life (t1/2)
Once you have the elimination rate constant, you can easily determine the biological half-life of the substance. The half-life is the time required for the concentration of the drug to decrease by 50%.
The relationship is defined by the equation:
t1/2 = 0.693 / ke
Example Calculation
Imagine a patient receives a dose of medication. Based on blood draws, you observe the following on the linear part of the semi-log graph:
At 2 hours (t1), the concentration is 50 mg/L (C1).
At 8 hours (t2), the concentration drops to 12.5 mg/L (C2).
Using the calculator above or the manual formula:
Time difference = 8 – 2 = 6 hours.
ln(50) ≈ 3.912
ln(12.5) ≈ 2.526
Difference in logs = 3.912 – 2.526 = 1.386
ke = 1.386 / 6 ≈ 0.231 hr⁻¹
This means approximately 23.1% of the remaining drug is eliminated every hour.