Reaction Rate Calculator — Rate Law, Integrated Rate Laws & Reaction Order
Calculate reaction rate using rate = k[A]^m[B]^n, determine the rate constant k and reaction orders using the method of initial rates, solve integrated rate laws for zero, first, and second order reactions, and identify reaction order from concentration-time data by linear regression — all with full step-by-step working. This reaction rate calculator and reaction rate constant calculator covers every standard kinetics problem.
Compute rate = k × [A]^m × [B]^n or find the rate constant k. Reaction orders must be determined experimentally.
Rate Law Calculation — rate = k × [A]^m × [B]^n
Enter initial rate experiment data. The method of initial rates compares experiments where one concentration varies to determine reaction orders. Minimum 2 experiments.
Method of Initial Rates — Rate Law Determination
Solve integrated rate laws for zero, first, and second order reactions. Find [A] at time t, the time to reach a target [A], or the half-life.
Integrated Rate Law Result
Enter concentration vs time experimental data. The calculator performs linear regression on all three linearized forms and identifies the reaction order from the best fit (highest R²).
| Point # | Time t (s) | [A] (M) |
|---|
Reaction Order Determination — Linear Regression Analysis
| Order | Differential | Integrated | Linear Plot | Slope | t½ |
|---|---|---|---|---|---|
| 0 | rate = k | [A] = [A]₀ − kt | [A] vs t | −k | [A]₀/(2k) |
| 1 | rate = k[A] | ln[A] = ln[A]₀ − kt | ln[A] vs t | −k | 0.693/k ★CONSTANT |
| 2 | rate = k[A]² | 1/[A] = 1/[A]₀ + kt | 1/[A] vs t | +k | 1/(k[A]₀) |
| Overall Order | k Units | Example | Note |
|---|---|---|---|
| 0 | M·s⁻¹ | Surface catalysis | rate = k |
| 1 | s⁻¹ | Radioactive decay | rate = k[A] |
| 2 | M⁻¹·s⁻¹ | Bimolecular collision | rate = k[A]² |
| 3 | M⁻²·s⁻¹ | Termolecular (rare) | rate = k[A]³ |
| Order | t½ Formula | Depends on [A]₀? | Constant? |
|---|---|---|---|
| 0 | [A]₀/(2k) | YES | NO — decreases over time |
| 1 | 0.693/k ★ | NO | YES — CONSTANT! |
| 2 | 1/(k[A]₀) | YES | NO — increases over time |
| If you have | Method |
|---|---|
| Two experiments, one [conc] varies | m = log(r₂/r₁) / log([A]₂/[A]₁) |
| [A] vs time data (multiple points) | Test 3 linearized plots; highest R² = order |
| Known elementary mechanism | Order = stoichiometric coefficient in elementary step |
| Rate doubles when [A] doubles | First order (m=1) |
| Rate quadruples when [A] doubles | Second order (m=2) |
| Rate unchanged when [A] changes | Zero order (m=0) |
| Plot | Linear = Order | Slope | y-intercept |
|---|---|---|---|
| [A] vs t | Zero order | −k | [A]₀ |
| ln[A] vs t | First order | −k | ln[A]₀ |
| 1/[A] vs t | Second order | +k (positive!) | 1/[A]₀ |
Rate Law — rate = k[A]^m[B]^n
The rate law (differential rate law) expresses how reaction rate depends on reactant concentrations: rate = k × [A]^m × [B]^n, where k is the rate constant, m and n are the reaction orders determined experimentally (NOT from stoichiometric coefficients, except for elementary reactions), and [A] and [B] are molar concentrations. The overall order equals m + n. This reaction rate constant calculator determines k and rate automatically. The differential rate law and rate law expression are the same thing — a rate expression relating rate to instantaneous concentrations.
The general form of a rate law requires that rate constant k is always positive for forward reactions — a negative k value is physically impossible for a simple forward reaction. The rate expression defines how rate equation changes with concentration at a given temperature. Elementary rate laws (for single-step reactions) have orders equal to stoichiometric coefficients, but complex reactions require experimental determination of the rate law.
How to solve for k in rate law: Rearrange to k = rate / ([A]^m × [B]^n). Always verify units: k units must make rate have units of M/s. Is rate constant always positive? Yes — k is always positive for forward reactions at any temperature.
Method of Initial Rates — Finding Reaction Order Experimentally
The method of initial rates uses initial rate data from experiments designed so only one concentration changes at a time. The reaction order for each reactant is found by the ratio method. Initial rate meaning: the rate measured at the very beginning of the reaction (t→0), before significant concentration changes occur.
How to determine order of reaction: If rate doubles when [A] doubles → m = log(2)/log(2) = 1 (first order). If rate quadruples → m = log(4)/log(2) = 2 (second order). If rate is unchanged → m = 0 (zero order). How to calculate order of reaction is always this logarithm ratio. How to find reaction order from table requires identifying which pairs of experiments hold all but one concentration constant.
Example: 2-Reactant Initial Rates Problem
- Experiments 1 and 2: [B] constant, [A] doubles → rate quadruples → m = log(4)/log(2) = 2
- Experiments 2 and 3: [A] constant, [B] doubles → rate doubles → n = log(2)/log(2) = 1
- Rate law: rate = k[A]²[B] — overall third order
- k = rate₁/([A]₁²×[B]₁) = 2.00×10⁻³/(0.10²×0.10) = 2.00 M⁻²s⁻¹
Zero Order Reactions — [A] = [A]₀ − kt
In a zero order reaction, the rate is constant and independent of concentration: rate = k. The integrated rate law for zero order is [A] = [A]₀ − kt — concentration decreases linearly with time. The zero order graph is [A] vs t: a straight line with slope = −k. Zero order reaction kinetics arise when a catalyst is saturated (enzyme at Vmax, surface catalysis at full coverage).
The zero order integrated rate law shows that half-life t½ = [A]₀/(2k) decreases as the reaction proceeds — each successive half-life is shorter because [A]₀ decreases. The zeroth order integrated rate law and 0 order integrated rate law are the same expression. Zero order reaction plot: [A] vs t gives a straight line. Zero order reaction graph: negative slope equal to −k.
Zero Order Example: Find time for [A] to drop from 0.50 M to 0.20 M, k=1.5×10⁻³ M/s
- t = ([A]₀ − [A])/k = (0.50 − 0.20)/1.5×10⁻³ = 0.30/0.00150
- t = 200 s
- t½ = 0.50/(2×0.00150) = 167 s (would be 0.30/0.00300 = 100s at [A]=0.30M — decreasing)
First Order Reactions — ln[A] = ln[A]₀ − kt
The first order integrated rate law gives [A] = [A]₀ × e^(−kt). The logarithmic form is ln[A] = ln[A]₀ − kt, which is linear in t. The most important property of first order kinetics: t½ = ln(2)/k = 0.6931/k is constant regardless of concentration. Every half-life, the concentration halves exactly. First order graph: ln[A] vs t is a straight line with slope = −k. The first order decay equation [A] = [A]₀·e^(−kt) appears in radioactive decay, drug metabolism, and many unimolecular reactions.
The first order kinetics equation is used extensively. First order integrated rate law formula: the integrated first order rate equation is [A]=[A]₀e^(−kt) or equivalently ln([A]/[A]₀) = −kt. Half life formula first order: t½ = 0.6931/k. This is the only order where the half-life is constant — independent of concentration. After 5 half-lives, only (1/2)⁵ = 3.125% of reactant remains. Half life of a first order reaction is what makes radioactive dating possible — the constant t½ is the unique signature.
First Order Example: [A]₀=0.800M, k=0.0200 s⁻¹, t=200s
- kt = 0.0200 × 200 = 4.00
- [A] = 0.800 × e^(−4.00) = 0.800 × 0.018316 = 0.01465 M
- Fraction remaining = 0.01465/0.800 = 1.83%
- t½ = 0.6931/0.0200 = 34.66 s (constant!)
- Number of half-lives elapsed: 200/34.66 = 5.77 half-lives
Second Order Reactions — 1/[A] = 1/[A]₀ + kt
The second order integrated rate law is 1/[A] = 1/[A]₀ + kt. The second order graph is 1/[A] vs t: a straight line with slope = +k (positive slope — crucial distinction from zero and first order which have negative slopes). Units of k for second order: M⁻¹s⁻¹. The half-life t½ = 1/(k[A]₀) increases as [A]₀ decreases — each successive half-life is longer. Second order kinetics arise from bimolecular collisions: 2A→products or A+B→products.
Second Order Example: [A]₀=0.200M, k=0.500 M⁻¹s⁻¹, t=100s
- 1/[A] = 1/0.200 + 0.500×100 = 5.00 + 50.0 = 55.0 M⁻¹
- [A] = 1/55.0 = 0.01818 M
- t½ = 1/(0.500×0.200) = 10.0 s (at this initial [A]₀)
- Second half-life (from [A]₀=0.100): t½ = 1/(0.500×0.100) = 20.0 s (longer!)
Identifying Reaction Order from Graphs
The graphical method for determining reaction order uses the same (t, [A]) data set transformed three ways. Whichever linearized plot gives the best straight line (highest R²) reveals the order. This is the standard kinetics graph method taught in every chemistry course.
- [A] vs t: straight line → zero order; slope = −k; y-intercept = [A]₀
- ln[A] vs t: straight line → first order; slope = −k; y-intercept = ln[A]₀
- 1/[A] vs t: straight line → second order; slope = +k; y-intercept = 1/[A]₀
The first order reaction graph shows exponential decay in [A] vs t, but converts to a straight line in the ln[A] vs t plot. The second order reaction graph shows a hyperbolic curve in [A] vs t, linear in 1/[A] vs t. The zero order reaction graph shows perfectly linear [A] vs t. How to determine order of reaction from graph: perform all three regressions; the one with R² closest to 1.000 is the correct order. How to find the rate constant from a graph: k = |slope| for zero and first order; k = slope for second order.
Key insight for reaction order graphs: All three plots are made from the SAME experimental data — you only transform the y-axis. The concentration vs time graph for first order reaction shows exponential decay, but the same data plotted as ln[A] vs t gives a perfect straight line. The slope of line for first order reaction equals −k, so k = 0.0200 s⁻¹ when slope = −0.0200 s⁻¹.
Rate Constant Units — How to Determine Units of k
The unit of rate constant k depends on the overall reaction order n using: units of k = M^(1−n)·s⁻¹. This formula ensures rate always has units of M/s = M¹·s⁻¹. Zero order: M^(1−0)·s⁻¹ = M·s⁻¹. First order: M^(1−1)·s⁻¹ = s⁻¹. Second order: M^(1−2)·s⁻¹ = M⁻¹·s⁻¹. Third order: M⁻²·s⁻¹. Calculate the rate constant with proper units by first determining the overall order, then applying this formula.
| Order | k units formula | k units | Equivalent |
|---|---|---|---|
| Zero | M^(1−0)·s⁻¹ | M·s⁻¹ | mol·L⁻¹·s⁻¹ |
| First | M^(1−1)·s⁻¹ | s⁻¹ | s⁻¹ |
| Second | M^(1−2)·s⁻¹ | M⁻¹·s⁻¹ | L·mol⁻¹·s⁻¹ |
| Third | M^(1−3)·s⁻¹ | M⁻²·s⁻¹ | L²·mol⁻²·s⁻¹ |
Can rate constant be negative? No — k is always positive for forward reactions. A negative k would violate thermodynamics. Is rate constant always positive? Yes at all temperatures. How do you calculate the rate constant k? Use k = rate/([A]^m×[B]^n) from any experimental data point after determining the orders.
Common Mistakes in Kinetics Calculations
Mistake 1 — Using Stoichiometry for Reaction Orders
- ❌ Wrong: For 2A→products, assume rate = k[A]²
- ✅ Correct: Orders come from experiment ONLY (unless told it's an elementary reaction)
- The rate law expression must be determined by the method of initial rates or by graphical analysis
Mistake 2 — Wrong k Units
- ❌ Wrong: Using k in s⁻¹ for a second order rate law
- ✅ Correct: Second order k must be in M⁻¹s⁻¹. Check: k[A]² = (M⁻¹s⁻¹)(M²) = M/s ✓
Mistake 3 — Sign Error in k from Graph Slope
- ❌ Wrong: k = slope of ln[A] vs t (slope is negative → k would be negative)
- ✅ Correct: k = −slope for zero and first order; k = +slope for second order
- First order: slope = −k = −0.020 → k = 0.020 s⁻¹ (always positive)
Mistake 4 — Half-Life Confusion
- ❌ Wrong: Assuming all reactions have constant t½
- ✅ Correct: ONLY first order has constant t½ = 0.693/k. Zero order t½ decreases; second order t½ increases as reaction proceeds.
Mistake 5 — Rate Law vs Integrated Rate Law Confusion
- ❌ Wrong: Using rate = k[A] to find [A] at time t
- ✅ Correct: Rate law gives instantaneous rate; integrated rate law gives [A] as function of t. Use [A] = [A]₀·e^(−kt) for first order.
Worked Examples — 8 Complete Kinetics Problems
1. Find Rate: rate = k[A]²[B], k=2.50×10⁻³ M⁻²s⁻¹, [A]=0.20M, [B]=0.15M
- rate = 2.50×10⁻³ × (0.20)² × (0.15)
- = 2.50×10⁻³ × 0.0400 × 0.150
- = 2.50×10⁻³ × 0.00600 = 1.50×10⁻⁵ M/s
2. Method of Initial Rates: 2-Reactant Table → Orders and k
- Exp 1,2: [B]=const, [A] doubles: rate ratio=4 → m=log(4)/log(2)=2
- Exp 2,3: [A]=const, [B] doubles: rate ratio=2 → n=log(2)/log(2)=1
- rate = k[A]²[B]; k = 2.00×10⁻³/((0.10)²×0.10) = 2.00 M⁻²s⁻¹
3. First Order Integrated Rate Law: [A]₀=0.800M, k=0.0200s⁻¹, t=200s
- [A] = 0.800 × e^(−0.0200×200) = 0.800 × e^(−4.00)
- = 0.800 × 0.01832 = 0.01465 M
- t½ = 0.6931/0.0200 = 34.66 s (constant)
4. Zero Order: Find Time for [A] 0.50M→0.20M, k=1.5×10⁻³ M/s
- t = ([A]₀−[A])/k = (0.50−0.20)/(1.5×10⁻³)
- t = 0.30/0.00150 = 200 s
5. Second Order: [A]₀=0.200M, k=0.500 M⁻¹s⁻¹, t=100s
- 1/[A] = 1/0.200 + 0.500×100 = 5.00 + 50.0 = 55.0 M⁻¹
- [A] = 1/55.0 = 0.01818 M
6. Half-Life: First Order k=0.0200s⁻¹; After 5 Half-Lives
- t½ = 0.6931/0.0200 = 34.66 s
- After 5 half-lives: fraction = (1/2)⁵ = 1/32 = 3.125% remains
- 5 half-lives elapsed at t = 5×34.66 = 173.3 s
7. Determine Order from Graphing — Which Plot is Linear?
- Compute [A], ln[A], 1/[A] for all time points
- Run linear regression on each: find slopes and R² values
- Best R² (closest to 1.0000) identifies the order — use Tool 4 above
8. Find k from Graph: Slope of ln[A] vs t = −0.0200 s⁻¹
- First order: ln[A] = ln[A]₀ − kt → slope = −k
- k = −(−0.0200) = 0.0200 s⁻¹
- Check units: k in s⁻¹ is correct for first order rate constant ✓