5.1 Reaction Rates

AP Chemistry · Unit 5 Kinetics

This is the starting point for all of kinetics: before we can ask why a reaction is fast or slow, we need a precise way to measure how fast it's going.

Learning Outcomes

Define reaction rate in terms of concentration change over time
Explain why the rate of disappearance of a reactant differs from the rate of appearance of a product (stoichiometry)
Write and interpret a rate expression for a general reaction
Calculate instantaneous rate from a concentration-vs-time graph (tangent slope)
Explain why the initial rate is the most commonly reported rate in kinetics experiments

Part 1: What Do We Mean by "Rate"?

Think about the word "rate." We use it all the time — heart rate, interest rate, data rate. A rate always tells you how much something changes per unit of time.

In chemistry, the "something" is concentration — usually in moles per liter, $\mathrm{mol/L}$, or molarity ($M$). And the time is in seconds, or minutes, depending on how fast the reaction is.

So reaction rate is: how fast does the concentration of something change?

$$\text{rate} = \frac{\Delta [\text{species}]}{\Delta t}$$

Simple enough. But there's a subtlety coming, so let's set up a concrete example first.

Part 2: A Concrete Reaction

Consider the decomposition of dinitrogen pentoxide:

$$2\,\mathrm{N_2O_5}(g) \longrightarrow 4\,\mathrm{NO_2}(g) + \mathrm{O_2}(g)$$

Two moles of $\mathrm{N_2O_5}$ break apart into four moles of $\mathrm{NO_2}$ and one mole of $\mathrm{O_2}$. Now I want to ask: what's the rate of this reaction?

[wait]

Here's the problem — the reactant is disappearing, and the products are appearing. And they're doing so at different speeds because the stoichiometry is different. For every 2 moles of $\mathrm{N_2O_5}$ that vanish, 4 moles of $\mathrm{NO_2}$ show up and 1 mole of $\mathrm{O_2}$ shows up.

So if I measure "how fast $\mathrm{N_2O_5}$ is disappearing," I get a different number than "how fast $\mathrm{O_2}$ is appearing." They're related, but they're not equal.

How do we reconcile this? We divide by the stoichiometric coefficient.

Part 3: The General Rate Expression

For a general reaction:

$$a\,A + b\,B \longrightarrow c\,C + d\,D$$

The unique reaction rate is defined as:

$$\text{rate} = -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = -\frac{1}{b}\frac{\Delta[B]}{\Delta t} = +\frac{1}{c}\frac{\Delta[C]}{\Delta t} = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$$

Three things to notice here:

Reactants get a negative sign. Their concentration is decreasing. We want the rate to be positive (a reaction rate is a magnitude), so we negate the negative change.

Products get a positive sign. Their concentration is increasing, so the change is already positive.

We divide by the stoichiometric coefficient. This normalizes everything — no matter which species you measure, you get the same rate for the reaction.

[pause]

Going back to our example:

$$\text{rate} = -\frac{1}{2}\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = +\frac{1}{4}\frac{\Delta[\mathrm{NO_2}]}{\Delta t} = +\frac{\Delta[\mathrm{O_2}]}{\Delta t}$$

So if $\mathrm{N_2O_5}$ is disappearing at $4 \times 10^{-3}\ \mathrm{M/s}$, then $\mathrm{NO_2}$ is appearing at $8 \times 10^{-3}\ \mathrm{M/s}$ (twice as fast, because the coefficient is 4 vs. 2), and $\mathrm{O_2}$ is appearing at $2 \times 10^{-3}\ \mathrm{M/s}$ (half as fast, because the coefficient is 1 vs. 2).

All three expressions give you the same rate: $2 \times 10^{-3}\ \mathrm{M/s}$.

$$\text{rate} = -\frac{1}{2}(-4 \times 10^{-3}) = \frac{1}{4}(8 \times 10^{-3}) = +1(2 \times 10^{-3}) = 2 \times 10^{-3}\ \mathrm{M/s}$$

That's the beauty of the stoichiometric normalization.

[pause]

Part 4: Average Rate vs. Instantaneous Rate

Now, just like in physics where we had average velocity and instantaneous velocity — here we have average rate and instantaneous rate.

The average rate is just total change over total time:

$$\bar{r} = \frac{\Delta[A]}{\Delta t}$$

But here's the thing about chemical reactions: the rate is usually fastest at the beginning, when concentrations are highest, and it slows down as reactants are consumed. So the average rate over, say, the first 100 seconds is different from the average rate over the next 100 seconds.

If I plot concentration vs. time:

[gesture — sketch a decreasing curve, flattening out]

The curve starts steep — that's high rate — and gradually flattens — that's low rate.

The instantaneous rate at any moment is the slope of the tangent line at that point on the curve. Same idea as in physics: slope = derivative.

$$r_{\text{inst}} = \frac{d[A]}{dt}$$

To find it from experimental data, you draw a tangent to the curve at the time you care about and calculate its slope.

[pause]

Part 5: Why Initial Rate?

When chemists report a reaction rate, they almost always report the initial rate — the rate at $t = 0$, the very beginning of the reaction.

Why? Because at $t = 0$:

The concentrations are known exactly (you just mixed the reagents)
No products have built up yet, so there's no reverse reaction to worry about
The slope of the tangent is steepest, so it's easiest to measure accurately

The initial rate is the cleanest, most reproducible number. And when we get to 5.2 — the Rate Law — we'll see that the initial rate is what lets us figure out how concentration affects rate. That's the whole basis of the method of initial rates.

[pause]

Part 6: Units

Quick but important: what are the units of reaction rate?

Concentration is in molarity, $\mathrm{mol/L}$. Time is in seconds. So:

$$\text{units of rate} = \mathrm{mol \cdot L^{-1} \cdot s^{-1}} = M/\mathrm{s}$$

Always. Regardless of the reaction. The rate always has units of concentration per time.

If a problem gives you minutes instead of seconds, convert. If it gives you pressure instead of concentration (for gas-phase reactions), you can work with pressure — but the AP exam expects molarity unless stated otherwise.

Worked Example

Problem: For the reaction

$$2\,\mathrm{N_2O_5}(g) \longrightarrow 4\,\mathrm{NO_2}(g) + \mathrm{O_2}(g)$$

the concentration of $\mathrm{N_2O_5}$ decreases from $1.00\ \mathrm{M}$ to $0.76\ \mathrm{M}$ in $40.0\ \mathrm{s}$. Calculate (a) the average rate of disappearance of $\mathrm{N_2O_5}$, (b) the average rate of appearance of $\mathrm{NO_2}$, and (c) the unique reaction rate.

Solution

(a) Rate of disappearance of $\mathrm{N_2O_5}$:

$$\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = \frac{0.76 - 1.00}{40.0} = \frac{-0.24}{40.0} = -6.0 \times 10^{-3}\ \mathrm{M/s}$$

The negative sign means it's disappearing. The rate of disappearance (as a positive magnitude) is $6.0 \times 10^{-3}\ \mathrm{M/s}$.

(b) Rate of appearance of $\mathrm{NO_2}$ — use the stoichiometric ratio. For every 2 moles of $\mathrm{N_2O_5}$ consumed, 4 moles of $\mathrm{NO_2}$ form:

$$\frac{\Delta[\mathrm{NO_2}]}{\Delta t} = -\frac{4}{2}\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = -2(-6.0 \times 10^{-3}) = 1.2 \times 10^{-2}\ \mathrm{M/s}$$

$$\text{rate} = -\frac{1}{2}\frac{\Delta[\mathrm{N_2O_5}]}{\Delta t} = -\frac{1}{2}(-6.0 \times 10^{-3}) = 3.0 \times 10^{-3}\ \mathrm{M/s}$$

Check: $\frac{1}{4}(1.2 \times 10^{-2}) = 3.0 \times 10^{-3}$. Consistent.

Common Mistakes

Forgetting the sign convention — reactants decrease (negative), products increase (positive). The overall rate should always be positive.
Forgetting to divide by the stoichiometric coefficient — each species gives a different raw rate; only dividing by the coefficient gives you the unique reaction rate.
Confusing average rate with instantaneous rate — the average rate over a long interval can be very different from the rate at a specific instant.
Not converting time units — if data is in minutes, convert to seconds before calculating rate in $\mathrm{M/s}$.

Summary

Reaction rate = change in concentration per unit time, $\mathrm{M/s}$
For $aA + bB \rightarrow cC + dD$: rate $= -\frac{1}{a}\frac{\Delta[A]}{\Delta t} = \cdots = +\frac{1}{d}\frac{\Delta[D]}{\Delta t}$
Reactants: negative sign (disappearing). Products: positive sign (appearing).
Instantaneous rate = slope of tangent on concentration-vs-time graph
Initial rate is the gold standard — cleanest measurement, no reverse reaction
Units: always $\mathrm{mol \cdot L^{-1} \cdot s^{-1}}$

Homework

For the reaction $\mathrm{H_2}(g) + \mathrm{I_2}(g) \longrightarrow 2\,\mathrm{HI}(g)$, the concentration of $\mathrm{H_2}$ decreases from $0.50\ \mathrm{M}$ to $0.38\ \mathrm{M}$ in $20\ \mathrm{s}$. Calculate the rate of disappearance of $\mathrm{H_2}$, the rate of appearance of $\mathrm{HI}$, and the unique reaction rate.
Why is the initial rate preferred over the average rate when studying reaction kinetics? Give two reasons.
Sketch a concentration-vs-time graph for a reactant in a reaction that starts fast and slows down. Mark where the instantaneous rate is highest and where it approaches zero.