First-Order Rate Constant: Formula and Example

The first-order rate constant is calculated from k = ln([A]₀/[A]_t)/t, where [A]₀ is the initial reactant concentration and [A]_t is its concentration after time t. This calculator uses those values to return k in the inverse of the time unit you enter, along with the reactant’s remaining fraction.

Formula

For a reaction that is first order in reactant A, the integrated rate law is ln([A]_t) = ln([A]₀) − kt. Rearranging gives k = ln([A]₀/[A]_t)/t. The equivalent common-log form is k = 2.303 log₁₀([A]₀/[A]_t)/t.

• k is the first-order rate constant, in inverse time, such as s⁻¹ when t is in seconds.
• [A]₀ is the initial concentration of A, in mol/L in this calculator.
• [A]_t is A’s concentration after elapsed time t, in the same concentration unit.
• t is the elapsed time, in seconds here. If time is entered in minutes, the result is min⁻¹ instead.

The concentration ratio has no units, so the logarithm is dimensionless. The rate constant’s unit comes from dividing by time. Because the logarithm uses a ratio, the two concentration values must use matching units; they do not have to be mol/L if both are expressed consistently.

Worked Example

Using the calculator’s default values, [A]₀ = 1 mol/L, [A]_t = 0.5 mol/L, and t = 60 s:

1. Find the concentration ratio: [A]₀/[A]_t = 1/0.5 = 2.
2. Take the natural logarithm: ln(2) = 0.6931471806.
3. Divide by elapsed time: k = 0.6931471806/60 s = 0.01155245301 s⁻¹, which rounds to 0.0116 s⁻¹ at four decimal places.
4. Calculate the remaining fraction: [A]_t/[A]₀ = 0.5/1 = 0.5000, or 50.00% remaining.

Assumptions

• The measured concentration decreases according to a single first-order rate law over the stated interval.
• Both concentration measurements refer to the same reactant and use the same units; elapsed time is positive.
• The rate constant is treated as constant during the interval. Changes in temperature or other reaction conditions can change its value.

What First-Order Means

A first-order reaction has a rate proportional to the concentration of A: rate = −d[A]/dt = k[A], for a reaction whose rate law is first order in A. The minus sign indicates that A is consumed. If the concentration is doubled while conditions remain the same, the instantaneous disappearance rate doubles. The integrated equation describes how concentration changes over a measured period, making it useful when you know concentrations at two times.

Reaction order is determined from experimental behavior or a justified mechanism; it cannot be inferred just from the balanced chemical equation in general. A reaction can also be first order in one reactant but depend on other reactants as well. In that case, the observed or pseudo-first-order constant may apply only while the other reactant concentrations are effectively constant.

How to Use and Interpret k

Enter the initial concentration, the concentration measured later, and the time between those measurements. The later concentration must be positive and lower than the initial concentration for a positive k under a consumption model. The calculator returns a numerical value and units based on seconds, plus the fraction remaining. A fraction of 0.5 means half the initial amount remains; it does not mean the rate constant is 0.5.

The rate constant is not the same as the reaction rate. The rate has concentration-per-time units and depends on concentration, while k has inverse-time units for a first-order law. For example, at a particular concentration, the disappearance rate is k[A]. A larger k means faster decay under the conditions represented, but values should only be compared when the reaction conditions and time units are compatible.

Half-Life and Useful Comparisons

For a first-order process, the half-life is the time required for concentration to fall to half its starting value. Setting [A]_t/[A]₀ = 1/2 in the integrated law gives t_1/2 = ln(2)/k. It does not depend on the starting concentration, provided the same first-order model and conditions apply. With the example k = 0.01155245301 s⁻¹, the half-life is approximately 60.000 s, consistent with the default values.

After each half-life, half of what remains is consumed: after two half-lives, one-quarter remains; after three, one-eighth remains. This differs from a zero-order process, whose concentration falls by a fixed amount per unit time, and a second-order process, whose integrated concentration relationship is different. A straight line in a plot of ln([A]) against time supports a first-order model; its slope is −k.

Common Errors and Limits

• Reversing the ratio: use [A]₀/[A]_t. Reversing it changes the sign of the logarithm.
• Mixing time units: a value calculated using minutes is in min⁻¹, not s⁻¹. Convert time before comparing constants.
• Using inconsistent concentration units: convert both values to a common unit. The ratio then has no units.
• Taking log base 10 without the factor: use ln directly, or multiply log₁₀ by 2.303.
• Overinterpreting rounded data: rounding measured concentrations can noticeably affect the logarithm, especially when the two values are close. Keep precision through intermediate calculations and report only precision justified by the measurements.

This two-point calculation produces a k value only if a first-order model is appropriate. Measurement noise, multiple reaction pathways, changing conditions, or concentrations near a detection limit can make a two-point estimate unreliable. For stronger evidence, calculate ln([A]) for multiple time points and inspect whether it follows a straight line; regression uses all points and can reveal deviations that two measurements conceal. If A increases, either the process is not a simple consumption case, the labels or measurements need checking, or the chosen model may not apply.

Frequently Asked Questions

What is the formula for a first-order rate constant?
Use k = ln([A]₀/[A]_t)/t. The concentration values must use the same units, and the unit of k is the inverse of the time unit.

What are the units of a first-order rate constant?
They are reciprocal time: s⁻¹ for time in seconds, min⁻¹ for time in minutes, or another corresponding inverse-time unit.

How do I calculate k from a half-life?
For a first-order reaction, k = ln(2)/t_1/2. Use the same time unit for the half-life and the resulting inverse-time unit for k.

Can I use log instead of ln?
Yes. With common logarithms, use k = 2.303 log₁₀([A]₀/[A]_t)/t. Omitting 2.303 gives an incorrect value.

Does a first-order reaction have a constant rate?
No. Its rate constant k is constant only while conditions remain fixed, but the reaction rate k[A] falls as the reactant concentration falls.

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in a0 | Initial concentration of A | mol/L | 1 | 0.000001
in at | Concentration of A at time t | mol/L | 0.5 | 0.000001
in t | Elapsed time | s | 60 | 0.000001
out k | First-order rate constant k | s^-1 | ln(a0 / at) / t | 4
out f | Fraction of A remaining | - | at / a0 | 4