What is the reaction rate?
Enter how a concentration changed over a time interval to get the average reaction rate, using the substance’s coefficient in the balanced equation. Or use the rate law rate = k[A]^m[B]^n to find the rate or the rate constant k with its units, and find a reaction order from two trials.
- Answer
- Rate = 0.0833333 mol/(L·h)
Rate = 0.0833333 mol/(L·h).
- Reaction rate
- 0.0833333
- Δ[ ] ÷ Δt
- −0.0833333
Answer: Rate = 0.0833333 mol/(L·h). Rate = 0.0833333 mol/(L·h).
How to calculate
Finds the average rate of a reaction from a concentration change over time and the stoichiometric coefficient, the rate or the rate constant k from the rate law rate = k[A]^m[B]^n, or a reaction order from two trials.
Example with the default inputs (Find Average rate, The substance is a Reactant, Its coefficient in the equation 1, Concentration at the start (mol/L) 1, Concentration at the end (mol/L) 0.5, Start time 0, End time 6, Time unit h): Rate = 0.0833333 mol/(L·h).
Method: Average rate = −(1/a) × Δ[reactant] ÷ Δt = (1/b) × Δ[product] ÷ Δt; rate = k[A]^m[B]^n; k = rate ÷ ([A]^m[B]^n), in L^(x−1) mol^(1−x) per time for x = m + n; order m = ln(rate₂ ÷ rate₁) ÷ ln([A]₂ ÷ [A]₁).
- Orders come from experiment: they need not match the coefficients of the balanced equation.
- In the order mode only [A] changes between the two trials; every other concentration and the temperature stay the same.
Worked examples
Each example is checked against the calculator on every build.
- Find Average rate, The substance is a Reactant, Its coefficient in the equation 1, Concentration at the start (mol/L) 1, Concentration at the end (mol/L) 0.5, Start time 0, End time 6, Time unit h gives Reaction rate 0.0833333, Δ[ ] ÷ Δt −0.0833333, Answer Rate = 0.0833333 mol/(L·h).Source: OpenStax, Chemistry 2e, §12.1 Chemical Reaction Rates (rate = −Δ[A]/Δt; for aA → bB, rate = −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt; H₂O₂ table), https://openstax.org/books/chemistry-2e/pages/12-1-chemical-reaction-rates, Figure 12.2 table: 0.0833 mol/(L·h)
- Find Average rate, The substance is a Reactant, Its coefficient in the equation 2, Concentration at the start (mol/L) 0.25, Concentration at the end (mol/L) 0.125, Start time 12, End time 18, Time unit h gives Reaction rate 0.0104167, Δ[ ] ÷ Δt −0.0208333.Source: OpenStax, Chemistry 2e, §12.1 Chemical Reaction Rates (rate = −Δ[A]/Δt; for aA → bB, rate = −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt; H₂O₂ table), https://openstax.org/books/chemistry-2e/pages/12-1-chemical-reaction-rates: Figure 12.2 gives −Δ[H₂O₂]/Δt = 0.0208 mol/(L·h)
- Find Rate constant, Rate (mol/(L·time)) 0.000066, [A] (mol/L) 0.000001, Order in A (m) 1, [B] (mol/L), optional 0.000003, Order in B (n) 1, Time unit s gives Rate constant k 22,000,000, Unit of k L/(mol·s), Answer k = 2.2 × 10⁷ L/(mol·s).Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.4 Determining a Rate Law from Initial Rates: k = 2.20 × 10⁷ L mol⁻¹ s⁻¹
- Find Rate constant, Rate (mol/(L·time)) 0.003, [A] (mol/L) 0.1, Order in A (m) 2, [B] (mol/L), optional 0.1, Order in B (n) 1, Time unit s gives Rate constant k 3, Unit of k L²/(mol²·s), Overall order 3.Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.5 Determining Rate Laws from Initial Rates: k = 3.0 mol⁻² L² s⁻¹
- Find Rate constant, Rate (mol/(L·time)) 0.00184, [A] (mol/L) 0.004, Order in A (m) 2, [B] (mol/L), optional 0.002, Order in B (n) 1, Time unit s gives Rate constant k 57,500.Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.5 Check Your Learning: k = 5.75 × 10⁴ mol⁻² L² s⁻¹
- Find Rate constant, Rate (mol/(L·time)) 0, [A] (mol/L) 0.00175, Order in A (m) 2, Order in B (n) 1, Time unit s gives Rate constant k 0.00000672653, Unit of k L/(mol·s).Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.4 Check Your Learning: k = 6.73 × 10⁻⁶ L/mol/s
- Find Rate law, Rate constant k 22,000,000, [A] (mol/L) 0.000002, Order in A (m) 1, [B] (mol/L), optional 0.000009, Order in B (n) 1, Time unit s gives Reaction rate 0.000396, Answer Rate = 3.96 × 10⁻⁴ mol/(L·s).Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.4: trial 4 rate 3.96 × 10⁻⁴ mol L⁻¹ s⁻¹
- Find Reaction order, Trial 1: [A] (mol/L) 0.1, Trial 1: rate 0.003, Trial 2: [A] (mol/L) 0.15, Trial 2: rate 0.00675 gives Reaction order 2, Answer Order = 2.Source: OpenStax, Chemistry 2e, §12.3 Rate Laws (rate = k[A]^m[B]^n; units of k), https://openstax.org/books/chemistry-2e/pages/12-3-rate-laws, Example 12.5: m = 2
How it works
Average rate (OpenStax §12.1). For a substance with coefficient a in the balanced equation, whose concentration goes from c₁ at time t₁ to c₂ at time t₂:
- Δ[ ] ÷ Δt = (c₂ − c₁) ÷ (t₂ − t₁)
- Reaction rate = −(1/a) × (c₂ − c₁) ÷ (t₂ − t₁) for a reactant, and +(1/a) × (c₂ − c₁) ÷ (t₂ − t₁) for a product.
With a coefficient of 1 the rate is the substance’s own rate of change (OpenStax’s table). The time unit (s, min or h) only names the unit: mol/(L·h) for hours.
Rate law (OpenStax §12.3). rate = k[A]^m[B]^n.
- Rate: k × [A]^m × [B]^n.
- Rate constant: k = rate ÷ ([A]^m × [B]^n).
Leave [B] empty for a rate law in A only (rate = k[A]^m); then n is not used. The orders m and n are any numbers from −3 to 3. The overall order is x = m + n (or m with no B), and k is in L^(x−1) mol^(1−x) per time unit, written:
| x − 1 | Unit of k (seconds) |
|---|---|
| 0 | s⁻¹ |
| 1 | L/(mol·s) |
| 2, 3, … | L²/(mol²·s), L³/(mol³·s), … |
| −1 | mol/(L·s) |
| −2, −3, … | mol²/(L²·s), mol³/(L³·s), … |
| not a whole number | (L/mol)^(x − 1)/s, such as (L/mol)^0.5/s |
An exponent written with a power of ten is in brackets: (L/mol)^(1 × 10⁻¹⁰)/s for x = 1.0000000001.
Reaction order from two trials. With only [A] changed between the trials: m = ln(rate₂ ÷ rate₁) ÷ ln([A]₂ ÷ [A]₁).
Exact values. Typed numbers are exact decimals. Powers with whole-number orders are exact; a fractional order’s power is the nearest double-precision number, worked on exactly from there. The logarithms in the order mode are double-precision numbers.
Rules the page checks.
- Average rate: concentrations from 0 to 100 mol/L, times from 0 to 10¹², the coefficient a whole number from 1 to 20. The end time must be after the start time. A reactant’s concentration must not rise and a product’s must not fall over the interval (the rate cannot be negative); otherwise no answer.
- Rate law: concentrations from 10⁻¹⁵ to 100 mol/L; k and rates from 10⁻¹⁰⁰ to 10¹⁰⁰.
- Order: concentrations from 10⁻¹⁵ to 100 mol/L and rates from 10⁻¹⁰⁰ to 10¹⁰⁰; the two concentrations must differ.
- An average rate, a Δ[ ] ÷ Δt or an overall order that is not 0 but smaller in size than 2⁻¹⁰²² (about 2.2 × 10⁻³⁰⁸), or too large for a number (above about 1.8 × 10³⁰⁸), has no answer.
How results are written. 6 significant figures, trailing zeros dropped, thousands commas. Rates and k are written m × 10ⁿ below 0.001 or from 1,000,000 (judged on the value before rounding) and plainly between; the order is written plainly (m × 10ⁿ only below 0.000001). The rows write a number below 0.000001 or from 10¹⁵ as m × 10ⁿ (an overall order of 1 × 10⁻⁷). The sentence reads “Rate = 0.0833333 mol/(L·h)”, “Rate = 3.96 × 10⁻⁴ mol/(L·s)”, “k = 2.2 × 10⁷ L/(mol·s)” or “Order = 2”.
Assumptions. Orders are experimental; in the order mode, nothing but [A] changes between the trials.
Worked examples by hand
H₂O₂, first 6 hours (OpenStax Figure 12.2). −(0.500 − 1.000) ÷ (6.00 − 0) = 0.0833333 mol/(L·h).
H₂O₂ from 12 h to 18 h, per mole of reaction. Δ[H₂O₂]/Δt = (0.125 − 0.250) ÷ 6 = −0.0208333; with a = 2, the rate = 0.0208333 ÷ 2 = 0.0104167 mol/(L·h).
NO + O₃ (Example 12.4). k = 6.60 × 10⁻⁵ ÷ (1.00 × 10⁻⁶ × 3.00 × 10⁻⁶) = 2.2 × 10⁷ L/(mol·s); from k, trial 4: 2.2 × 10⁷ × 2.00 × 10⁻⁶ × 9.00 × 10⁻⁶ = 3.96 × 10⁻⁴ mol/(L·s).
2NO + Cl₂ (Example 12.5). k = 0.00300 ÷ (0.10² × 0.10) = 3 L²/(mol²·s); order in NO from trials 1 and 3: ln(0.00675 ÷ 0.00300) ÷ ln(0.15 ÷ 0.10) = ln 2.25 ÷ ln 1.5 = 2.
OCl⁻ + I⁻ (Check Your Learning). k = 0.00184 ÷ (0.0040² × 0.0020) = 57,500 L²/(mol²·s).
CH₃CHO (Check Your Learning). k = 2.06 × 10⁻¹¹ ÷ (1.75 × 10⁻³)² = 6.72653 × 10⁻⁶ L/(mol·s) (OpenStax: 6.73 × 10⁻⁶).
Other questions people ask
How do I calculate the average rate of a reaction?
Divide the change in concentration by the time it took, and make it positive: rate = −Δ[A]/Δt for a reactant. H₂O₂ falls from 1.000 to 0.500 mol/L in 6.00 h, so its rate of decrease is 0.500 ÷ 6.00 = 0.0833 mol/(L·h) (OpenStax).
Why divide by the coefficient?
So that every substance gives the same rate for the reaction. For aA → bB, rate = −(1/a)Δ[A]/Δt = (1/b)Δ[B]/Δt. In 2H₂O₂ → 2H₂O + O₂, H₂O₂ disappears twice as fast as O₂ appears, so the reaction rate is half of H₂O₂’s rate of decrease.
How do I calculate the rate constant?
Divide the rate by the concentration terms of the rate law: k = rate ÷ ([A]^m[B]^n). For NO + O₃ with rate = k[NO][O₃]: k = 6.60 × 10⁻⁵ ÷ (1.00 × 10⁻⁶ × 3.00 × 10⁻⁶) = 2.20 × 10⁷ L/(mol·s).
What are the units of the rate constant?
They depend on the overall order x = m + n: mol/(L·s) for zero order, s⁻¹ for first order, L/(mol·s) for second order, L²/(mol²·s) for third order; in general L^(x−1) mol^(1−x) s⁻¹.
How do I find the order of a reaction from data?
Compare two trials where only one concentration changes: m = ln(rate₂ ÷ rate₁) ÷ ln([A]₂ ÷ [A]₁). Raising [NO] from 0.10 to 0.15 M raised the rate from 0.00300 to 0.00675: ln 2.25 ÷ ln 1.5 = 2, so the reaction is second order in NO.
Do the orders come from the balanced equation?
No. Orders are found by experiment and need not match the coefficients; they are usually 0, 1 or 2 but can be fractions or negative (OpenStax). The rate law of 2NO + Cl₂ → 2NOCl happens to be rate = k[NO]²[Cl₂].