Rate of reaction — average vs instantaneous

Overview

The rate of a chemical reaction measures how fast reactants convert to products. We need two different definitions because reactions don't proceed at constant speed — they usually slow down as reactants get used up. Average rate tells you the overall speed over a time interval, while instantaneous rate captures the exact speed at one specific moment.

Core Concepts

The negative sign converts the decrease in [A] to a positive rate (we report rates as positive numbers).

For a product P being formed: $r_{\text{avg}} = +\frac{\Delta[\text{P}]}{\Delta t} = \frac{[\text{P}]_2 - [\text{P}]_1}{t_2 - t_1}$

WHY this definition? Rate means "how much concentration changes per unit time." Δ[A]/Δt is the slope of the concentration-time graph over the interval. The negative sign is a sign convention: chemists want "rate" to be positive even though [reactant] drops.

WHY? As Δt shrinks to zero, the average rate over that tiny interval becomes the exact rate at a single time point. Mathematically, this is the derivative — the slope of the tangent line to the concentration curve at time t.

$\frac{-1}{a}\frac{d[\text{A}]}{dt} = \frac{-1}{b}\frac{d[\text{B}]}{dt} = \frac{1}{c}\frac{d[\text{C}]}{dt} = \frac{1}{d}\frac{d[\text{D}]}{dt} = r$

Derivation from stoichiometry:

1. When $a$ moles of A disappear, $c$ moles of C form (stoichiometry).
2. So if A's concentration drops by $a$ mol/L in time $\Delta t$, C's concentration rises by $c$ mol/L.
3. Rate per mole of A = $-\frac{1}{a}\frac{d[\text{A}]}{dt}$; rate per mole of C = $\frac{1}{c}\frac{d[\text{C}]}{dt}$.
4. These must be equal because they describe the same reaction event happening at the same speed.

WHY divide by stoichiometric coefficients? A balanced equation might consume2 moles of A for every 1 mole of B. Then [A] drops twice as fast as [B], but the reaction proceeds at one rate. Dividing by coefficients normalizes to the "reaction rate."

Worked Examples

Solution:

Step 1: Average rate of N₂O₅ consumption. $-\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = -\frac{0.0250 - 0.0400}{100 - 0} = -\frac{-0.0150}{100} = 1.50 \times 10^{-4} \text{ M/s}$ Why this step? We apply the definition directly. The negative of a negative gives positive rate.

Step 2: Convert to the reaction rate using stoichiometry. $r = \frac{-1}{2}\frac{\Delta[\text{N}_2\text{O}_5]}{\Delta t} = \frac{1}{2} \times 1.50 \times 10^{-4} = 7.50 \times 10^{-5} \text{ M/s}$ Why divide by 2? The coefficient of N₂O₅ is 2. The reaction "happens" at half the speed that N₂O₅ molecules disappear.

Step 3: Rate of NO₂ formation. $\frac{1}{4}\frac{d[\text{NO}_2]}{dt} = r \implies \frac{d[\text{NO}_2]}{dt} = 4r = 4 \times 7.50 \times 10^{-5} = 3.00 \times 10^{-4} \text{ M/s}$ Why multiply by 4? For every "reaction event," 4 NO₂ molecules form (stoichiometry). So NO₂ appears 4 times faster than the normalized reaction rate.

Answer: Reaction rate = $7.50 \times 10^{-5}$ M/s; NO₂ formation rate = $3.00 \times 10^{-4}$ M/s.

Solution:

Step 1: The tangent slope IS the derivative $\frac{d[\text{A}]}{dt}$ at t=50 s. $\frac{d[\text{A}]}{dt} = -0.0030 \text{ M/s}$ Why? Definition of derivative: slope of tangent.

Step 2: Apply the negative sign to get positive rate. $r_{\text{inst}} = -\frac{d[\text{A}]}{dt} = -(-0.0030) = 0.0030 \text{ M/s}$ Why? Sign convention: rate is positive even though [A] decreases.

Answer: 0.0030 M/s at t=50 s.

Solution:

Step 1: Average rate (consumption of H₂O₂). $-\frac{\Delta[\text{H}_2\text{O}_2]}{\Delta t} = -\frac{0.50 - 1.00}{1000 - 0} = 5.0 \times 10^{-4} \text{ M/s}$ Why? Definition of average rate.

Step 2: Reaction rate (account for stoichiometry). $r_{\text{avg}} = \frac{1}{2} \times 5.0 \times 10^{-4} = 2.5 \times 10^{-4} \text{ M/s}$ Why divide by 2? Coefficient is 2.

Step 3: Initial instantaneous rate (given directly). $r_{\text{inst}}(t=0) = \frac{1}{2} \times 1.5 \times 10^{-3} = 7.5 \times 10^{-4} \text{ M/s}$ Wait, why divide? The problem states "instantaneous rate" = 1.5×10⁻³ M/s. If that's the consumption rate of H₂O₂, we divide by 2 for reaction rate. (Clarify units in real problems!)

Comparison: $r_{\text{inst}}(t=0) = 7.5 \times 10^{-4}$ M/s is 3 times larger than $r_{\text{avg}} = 2.5 \times 10^{-4}$ M/s.

WHY? Reactions typically slow down because [reactant] decreases. The initial rate (when [H₂O₂] is highest) is faster than the average over1000 s (which includes later slow periods).

Why this feels right: You calculated the slope directly and got a negative number.

The fix: Rates are defined as positive quantities. For a disappearing reactant, you MUST include the negative sign: $r = -\frac{\Delta[\text{A}]}{\Delta t} = -(-0.03) = +0.03$ M/s. The negative in the definition compensates for the negative slope.

Steel-man: The confusion arises because in math, slopes can be negative. But in chemistry, "rate" is a speed concept (always positive), and we choose the sign in the definition to ensure this.

Why this feels right: You measured how fast A disappears, that's the rate, right?

The fix: The reaction rate is defined per reaction event, not per molecule. $r = \frac{-1}{2}\frac{d[\text{A}]}{dt} = \frac{0.04}{2} = 0.02 \text{ M/s}$ Why? Two A molecules vanish per reaction, so the reaction happens half as often as A molecules disappear.

Steel-man: The confusion comes from everyday language: "the rate A disappears" sounds like "the rate of reaction." But chemists need a single number that's the same no matter which species you look at. That's why we normalize by stoichiometry.

Why this feels right: 50 s is in the middle of the interval, so it seems like a good estimate.

The fix: Δ[A]/Δt is the average over 0-100 s. The instantaneous rate at t=50 s requires the tangent slope at exactly t=50 s, which could be very different if the reaction is speding up or slowing down.

When is the approximation okay? If the reaction rate is nearly constant (rare!), or if Δt is very small, average≈ instantaneous.

Memory Aids

Active Recall Practice

#flashcards/chemistry

What is the formula for average rate of reaction in terms of reactant A? :: $r_{\text{avg}} = -\frac{\Delta[\text{A}]}{\Delta t} = -\frac{[\text{A}]_2 - [\text{A}]_1}{t_2 - t_1}$ (negative sign converts decrease to positive rate)

Connections

• Collision theory — explains WHY rates decrease (fewer reactant molecules → fewer collisions)
• Rate law and order of reaction — shows how instantaneous rate depends on [reactant]
• Integrated rate laws — converts instantaneous rate expressions into [A] vs. t equations
• Half-life of reactions — uses instantaneous rate concept to define characteristic time
• Reaction mechanisms — multi-step reactions have rate-determining step (instantaneous rate concept)
• Catalysts — change the instantaneous rate without appearing in stoichiometry
• Arrhenius equation — temperature dependence of the rate constant (affects instantaneous rate)

Concept Map

Hinglish (regional understanding)