Comprehensive Guide to First-Order Reaction Kinetics
In chemical kinetics, a first-order reaction is a chemical transformation in which the reaction rate depends linearly on the concentration of a single reactant. When the reactant concentration doubles, the instantaneous velocity of the transformation doubles as well. Unlike zero-order reactions, where the reaction speed remains independent of how much reactant remains, or higher-order pathways like those computed with our Second-Order Reaction Calculator, first-order processes proceed via clean, predictable exponential decay.
First-order rate kinetics govern diverse phenomena across physical science and engineering. These include radioactive disintegration, thermal cracking of petrochemical hydrocarbons, atmospheric photolysis, thermal isomerization of organic complexes, and pharmacokinetics (in which drug clearance from human plasma follows first-order metabolism). Understanding the mathematical relationships behind first-order systems allows chemists to design batch reactors, establish shelf-life expiration dates for pharmaceuticals, and determine underlying reaction pathways.
Mathematical Derivations of Integrated Rate Laws
Consider an elementary unimolecular decomposition reaction where reactant A converts into products:
The differential rate law for this transformation is expressed as the rate of disappearance of reactant A over time:
Where:
- [A]: The instantaneous molar concentration of reactant A at any given time t.
- d[A] / dt: The rate of change of concentration with respect to time.
- k: The first-order rate constant, carrying fundamental dimensional units of reciprocal time (s⁻¹, min⁻¹, h⁻¹, or day⁻¹).
- t: The reaction duration or elapsed time.
Separation of Variables and Integration
To determine how concentration varies as a function of elapsed time, separate the variables concentration [A] and time t:
Integrating both sides from initial boundary conditions (t = 0, where [A] = [A]₀) to a future time t (where [A] = [A]):
Evaluating these standard integrals yields the natural logarithmic integrated rate equation:
Rearranging into the standard linear form (y = m * x + b):
This formulation shows that plotting the natural logarithm of concentration ln([A]) on the y-axis against time t on the x-axis yields a straight line with a slope equal to -k and a vertical y-intercept equal to ln([A]₀). Deviations from linearity in such a plot indicate that the process does not obey first-order kinetics.
Exponential Form of the Integrated Rate Law
Taking the exponential of both sides yields the direct concentration equation:
This exponential form demonstrates that the remaining reactant concentration decreases exponentially toward zero, but theoretically never reaches absolute zero in finite time. The fraction remaining, [A] / [A]₀, depends solely on the product of the rate constant and time (k * t), and is independent of the initial starting quantity.
The Chemistry of Reaction Half-Life (t1/2)
The half-life (t1/2) of a reaction is defined as the time required for the reactant concentration to decrease to exactly half of its initial value, such that [A] = 0.5 * [A]₀. Substituting this condition into the integrated rate law:
ln(0.5) = -k * t1/2
-0.693147 = -k * t1/2
Solving directly for the half-life yields:
This result highlights a hallmark of first-order kinetics: the half-life is entirely independent of initial reactant concentration [A]₀. Whether starting with 10.0 M, 1.0 M, or 0.001 M, the time required to consume 50% of the remaining reactant remains constant. This stands in contrast to zero-order reactions (where half-life is directly proportional to initial concentration) and second-order reactions (where half-life is inversely proportional to initial concentration).
This concentration-independent behavior is why first-order kinetics are used to model radioactive isotopes, which can be explored in detail using our Half-Life Calculator and Radioactive Decay Calculator.
Temperature Dependence: Connecting to the Arrhenius Equation
While the first-order rate constant k is independent of reactant concentration, it is sensitive to temperature. The temperature dependence of k is governed by the Arrhenius equation:
Where:
- A: The pre-exponential frequency factor (with identical units to k, s⁻¹).
- Ea: The activation energy barrier of the reaction in J/mol or kJ/mol.
- R: The universal gas constant (8.31446 J/(mol*K)).
- T: Absolute thermodynamic temperature in Kelvin (K).
As temperature increases, the proportion of molecular collisions possessing kinetic energy equal to or exceeding the activation energy threshold increases exponentially. Evaluating the magnitude of this kinetic barrier is straightforward with our Activation Energy Calculator and Arrhenius Equation Calculator.
Step-by-Step Practical Calculation Examples
Worked Example 1: Thermal Decomposition of Dinitrogen Pentoxide
The gas-phase decomposition of dinitrogen pentoxide (2 N₂O₅ -> 4 NO₂ + O₂) follows first-order kinetics with a rate constant of k = 0.00050 s⁻¹ at 45 °C. If a sealed reaction vessel initially contains 0.250 M of N₂O₅, determine the remaining concentration after 1,200 seconds of reaction.
- Identify Parameters:
[A]₀ = 0.250 M
k = 0.00050 s⁻¹
t = 1200 s - Compute the Exponent (-k * t):
-k * t = -(0.00050 s⁻¹) * (1200 s) = -0.600 - Calculate the Exponential Factor:
exp(-0.600) = 0.54881 - Determine Final Concentration:
[A] = [A]₀ * exp(-k * t) = 0.250 M * 0.54881 = 0.1372 M - Calculate Reactant Consumption Percentage:
Reacted = [A]₀ - [A] = 0.250 - 0.1372 = 0.1128 M
Conversion = (0.1128 / 0.250) * 100 = 45.12%
Conclusion: After 20 minutes (1,200 seconds), 0.1372 M of N₂O₅ remains unreacted, corresponding to a 45.12% conversion to nitrogen dioxide and oxygen.
Worked Example 2: Determining Required Reaction Time for 99% Decomposition
An industrial sterilization process degrades an unstable peroxide intermediate with a verified first-order rate constant of k = 0.035 min⁻¹. How long must the batch remain in the reactor to achieve 99.0% decomposition of the peroxide?
- Determine Remaining Fraction:
If 99.0% has decomposed, the remaining fraction is:
[A] / [A]₀ = 1.00 - 0.990 = 0.010 (or 1.0% remains) - Rearrange the Integrated Rate Law for Time:
t = -ln([A] / [A]₀) / k - Calculate the Natural Logarithm:
ln(0.010) = -4.60517 - Solve for Time (t):
t = -(-4.60517) / 0.035 min⁻¹ = 4.60517 / 0.035 = 131.58 minutes - Verify with Half-Lives:
t1/2 = 0.69315 / 0.035 = 19.80 minutes
Number of half-lives required = 131.58 / 19.80 = 6.64 half-lives
(0.5)^6.64 = 0.0100 (matches 99% decomposition)
Conclusion: The reactor must operate for approximately 131.6 minutes (2 hours, 11 minutes, and 35 seconds) to achieve the target 99.0% degradation.
Comparison of Reaction Orders
The table below summarizes the contrasting mathematical and kinetic characteristics of common elementary reaction orders:
| Order | Differential Rate Law | Integrated Rate Law | Linear Plot Axis (y vs. x) | Half-Life Formula | Units of k |
|---|---|---|---|---|---|
| Zero-Order | Rate = k | [A] = [A]₀ - k*t | [A] vs. t | t1/2 = [A]₀ / (2*k) | M*s⁻¹ |
| First-Order | Rate = k*[A] | ln([A]) = ln([A]₀) - k*t | ln([A]) vs. t | t1/2 = 0.693 / k | s⁻¹ |
| Second-Order | Rate = k*[A]² | 1/[A] = 1/[A]₀ + k*t | 1/[A] vs. t | t1/2 = 1 / (k*[A]₀) | M⁻¹*s⁻¹ |
Frequently Asked Questions (FAQ)
Rate = k * [A], substituting units gives M/s = [units of k] * M. Dividing both sides by M isolates the units of k as 1/s (or s⁻¹). Because concentration cancels out, the numerical value of k is independent of the volume or concentration units used.ln([A]/[A]₀) = -k*t cancels [A]₀, leaving t1/2 = ln(2)/k = 0.693/k.ln([A]) versus time will produce a straight line. The negative slope of this line corresponds directly to the rate constant (-k).Rate = k*[Ester]*[H2O] to Rate = k_obs*[Ester], where k_obs = k*[H2O].Delta[A] / Delta t. Because the reaction continuously slows down as reactant is consumed, the average rate changes over any given window. The instantaneous rate is the rate at a specific point in time, given by the derivative -d[A]/dt = k*[A], which corresponds to the tangent slope of the concentration-versus-time curve at that moment.ln(P) = ln(P0) - k*t.