Back to homeAP® Chemistry · AP Chemistry · Change subject
Start for free
AP Chem · AP® Chemistry
Chemical transformations · Rate and mechanisms

Reaction Rates, Collisions, and Activation Energy

A steel wool pad can sit in a drawer for months and barely rust, but hold it in a flame and it glows and burns in seconds. The chemical change is the same: iron reacting with oxygen. What changed is how fast it happens. Chemical kinetics is the study of how fast reactions go and why. By the end of this page you will be able to explain why finely divided iron burns so much faster than a solid bar.

Average and instantaneous rate

Drop a piece of magnesium ribbon into hydrochloric acid. At first the bubbles stream off. A few minutes later they are barely visible. How can we put a number on a rate that changes every moment?

Concentration of A versus time: a dashed secant line between t₁ and t₂ gives the average rate, and the solid tangent line at t₁ gives the instantaneous rate. [A]t t₁t₂ Secant: average rate · Tangent: instantaneous rate

The real curve of $[\text{A}]$ versus time is not a straight line. Early on there is plenty of reactant, so the rate is high. As A is used up, the rate drops. The instantaneous rate of disappearance is the negative of the slope of the tangent line to the $[\text{A}]$ vs. $t$ curve. Divide by the coefficient of A to get the rate of the reaction.

Practice★☆☆Not started
In a closed reactor of constant volume at $300\,\text{K}$, the gas-phase decomposition of dinitrogen pentoxide is monitored according to the reaction:…

In a closed reactor of constant volume at $300\,\text{K}$, the gas-phase decomposition of dinitrogen pentoxide is monitored according to the reaction:

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

Over an interval of $40\,\text{s}$, the concentration of $\mathrm{N_2O_5}$ decreases from $0.100\,\text{mol}\times\text{L}^{-1}$ to $0.064\,\text{mol}\times\text{L}^{-1}$.

  1. a) Calculate the average rate of disappearance of $\mathrm{N_2O_5}$ over that time interval.
  2. b) Determine the overall average rate of the reaction normalized by the stoichiometry, and the average rate of formation of $\mathrm{NO_2}$.
Solve
Practice★★☆Not started
The decomposition of reactant $\mathrm{A}$ in aqueous solution according to $\mathrm{A}\,(\mathrm{aq}) \rightarrow \mathrm{B}\,(\mathrm{aq}) + \mathrm{C}\,(\mathrm{aq})$…

The decomposition of reactant $\mathrm{A}$ in aqueous solution according to $\mathrm{A}\,(\mathrm{aq}) \rightarrow \mathrm{B}\,(\mathrm{aq}) + \mathrm{C}\,(\mathrm{aq})$ at $25\,^\circ\text{C}$ is studied kinetically. The concentrations of $\mathrm{A}$ are measured as a function of time, giving: at $t = 0\,\text{s}$, $[\mathrm{A}] = 0.80\,\text{mol}\times\text{L}^{-1}$; at $t = 10\,\text{s}$, $[\mathrm{A}] = 0.50\,\text{mol}\times\text{L}^{-1}$; at $t = 20\,\text{s}$, $[\mathrm{A}] = 0.32\,\text{mol}\times\text{L}^{-1}$; and at $t = 30\,\text{s}$, $[\mathrm{A}] = 0.22\,\text{mol}\times\text{L}^{-1}$. The observed variation over time is illustrated below.

0 10 20 30 0.20 0.40 0.60 0.80 t (s) [A] (mol/L) (0; 0.80) (10; 0.50) (20; 0.32) (30; 0.22)
  1. a) Calculate the mean rate of disappearance of $\mathrm{A}$ over the three successive intervals: $[0, 10]\,\text{s}$, $[10, 20]\,\text{s}$ and $[20, 30]\,\text{s}$.
  2. b) Compare the three values numerically and explain, using collision theory, the cause of the variation observed as the reaction proceeds.
Solve
Practice★★☆Not started
For the thermal decomposition reaction $2\,\mathrm{NO_2}\,(\mathrm{g}) \rightarrow 2\,\mathrm{NO}\,(\mathrm{g}) + \mathrm{O_2}\,(\mathrm{g})$ at constant volume and…

For the thermal decomposition reaction $2\,\mathrm{NO_2}\,(\mathrm{g}) \rightarrow 2\,\mathrm{NO}\,(\mathrm{g}) + \mathrm{O_2}\,(\mathrm{g})$ at constant volume and $500\,\text{K}$, the concentration of nitrogen dioxide is recorded against time. At time $t = 40\,\text{s}$, the tangent drawn to the concentration–time curve passes exactly through the auxiliary points $(0\,\text{s};\,0.052\,\text{mol}\times\text{L}^{-1})$ and $(80\,\text{s};\,0.012\,\text{mol}\times\text{L}^{-1})$.

  1. a) Determine the value of the instantaneous rate of disappearance of $\mathrm{NO_2}$ at that time $t = 40\,\text{s}$.
  2. b) Calculate the instantaneous rate of formation of $\mathrm{O_2}$ at that same time and justify why the instantaneous rate provides more precise kinetic information than the mean rate calculated between $0\,\text{s}$ and $80\,\text{s}$.
Solve

The collision model

In one liter of gas at room temperature, more than $10^{28}$ molecular collisions happen every second. Yet most reactions do not finish instantly. Why do nearly all collisions fail?

Two collisions between CO and NO₂: one with a favorable orientation that can lead to products, and one with an unfavorable orientation that bounces apart. CO NO₂ favorable orientation C end of COhits O of NO₂ CO NO₂ unfavorable

Only a tiny fraction of collisions meet both requirements. That is why reactions are not instantaneous even though collisions are constant.

Why temperature matters

At any temperature, molecules have a spread of kinetic energies, shown by the Maxwell-Boltzmann distribution. Only the molecules in the high-energy tail have energy $\geq E_a$. Raising the temperature shifts the whole curve to higher energies and flattens it, so a much larger fraction of collisions is above $E_a$. The rate rises sharply, usually much more than the small increase in collision frequency would suggest.

Maxwell-Boltzmann distributions at a lower and a higher temperature with a vertical line at the activation energy; the area to the right of the line is larger at the higher temperature. Number of molecules Kinetic energy Eₐ lower T higher T
Practice★☆☆Not started
Consider the overall gas-phase reaction between carbon monoxide and nitrogen dioxide: $$\mathrm{CO}\,(\mathrm{g}) + \mathrm{NO_2}\,(\mathrm{g}) \rightarrow…

Consider the overall gas-phase reaction between carbon monoxide and nitrogen dioxide:

$$\mathrm{CO}\,(\mathrm{g}) + \mathrm{NO_2}\,(\mathrm{g}) \rightarrow \mathrm{CO_2}\,(\mathrm{g}) + \mathrm{NO}\,(\mathrm{g})$$

Use a simplified collision model with transfer of an oxygen atom between these molecules, without necessarily identifying it with the complete real mechanism:

  1. a) Explain why not all collisions between $\mathrm{CO}$ and $\mathrm{NO_2}$ molecules lead to the formation of products, stating the two essential requirements for a collision to be effective.
  2. b) Justify which of the two relative orientations favors the reaction: the carbon atom of $\mathrm{CO}$ striking an oxygen atom of $\mathrm{NO_2}$, or the oxygen atom of $\mathrm{CO}$ striking the nitrogen atom of $\mathrm{NO_2}$.
Solve
Practice★☆☆Not started
In the industrial synthesis of ammonia: $$\mathrm{N_2}\,(\mathrm{g}) + 3\,\mathrm{H_2}\,(\mathrm{g}) \rightleftharpoons 2\,\mathrm{NH_3}\,(\mathrm{g}) \quad \Delta…

In the industrial synthesis of ammonia:

$$\mathrm{N_2}\,(\mathrm{g}) + 3\,\mathrm{H_2}\,(\mathrm{g}) \rightleftharpoons 2\,\mathrm{NH_3}\,(\mathrm{g}) \quad \Delta H^\circ = -92.2\,\text{kJ}$$
  1. a) Distinguish conceptually between the rate of a chemical reaction and its extent or equilibrium yield.
  2. b) If the working temperature is increased from $300\,^\circ\text{C}$ to $500\,^\circ\text{C}$, justify which kinetic parameter increases and what happens to the thermodynamic equilibrium constant $K_p$, explaining why a higher rate does not imply a greater final amount of product.
Solve

Energy profiles and activation energy

Play with the widget. Raising $E_a$ makes the hill taller (slower reaction). Changing $\Delta H$ only moves the final level relative to the starting level and does not change how high the peak is.

A catalyst provides a different pathway with a lower activation energy. It lowers the barrier in both directions by the same amount, so it speeds up the reaction without changing $\Delta H$.

Multistep profiles

Many reactions occur in several elementary steps. Their profile has one peak per step, and between peaks lies a valley. Each valley is an intermediate. The barrier of each step is measured from the energy of that step's own reactant up to its transition state, so $E_a$ (step) = $E_{TS} - E_{\text{reactant of that step}}$. For step 2, the reactant is the intermediate. The step with the largest barrier measured this way is usually the slowest, the rate-determining step. It is not always the one with the highest peak on the diagram. The overall $\Delta H$ is still just products minus reactants.

Two-step energy profile with two peaks and an intermediate in the valley between them; the second peak is higher, so the second step is rate-determining. Potential energy Reaction progress TS 1 TS 2 intermediate reactants products
Practice★★☆Not started
The potential energy diagram for the elementary gas-phase process: $$\mathrm{A}\,(\mathrm{g}) + \mathrm{B}\,(\mathrm{g}) \rightleftharpoons \mathrm{C}\,(\mathrm{g}) +…

The potential energy diagram for the elementary gas-phase process:

$$\mathrm{A}\,(\mathrm{g}) + \mathrm{B}\,(\mathrm{g}) \rightleftharpoons \mathrm{C}\,(\mathrm{g}) + \mathrm{D}\,(\mathrm{g})$$

shows that the enthalpy of the reactants is $H_R = 45\,\text{kJ}\times\text{mol}^{-1}$, the enthalpy of the products is $H_P = 85\,\text{kJ}\times\text{mol}^{-1}$, and the energy corresponding to the transition state (activated complex) is $H_{\neq} = 130\,\text{kJ}\times\text{mol}^{-1}$.

  1. a) Calculate the enthalpy of the forward reaction ($\Delta H^\circ$), the forward activation energy ($E_{a,\text{dir}}$) and the activation energy of the reverse process ($E_{a,\text{inv}}$).
  2. b) Explain what the activated complex represents physically and justify which of the two processes (forward or reverse) will have the greater rate constant at the same temperature, assuming identical pre-exponential factors.
Solve
Practice★★★Not started
A gas-phase reaction proceeds along the uncatalyzed forward energy profile with $E_{a} = 115.0\,\text{kJ}\times\text{mol}^{-1}$ and standard enthalpy change…

A gas-phase reaction proceeds along the uncatalyzed forward energy profile with $E_{a} = 115.0\,\text{kJ}\times\text{mol}^{-1}$ and standard enthalpy change $\Delta H^\circ = -35.0\,\text{kJ}\times\text{mol}^{-1}$. When a solid heterogeneous catalyst is added, the forward activation energy is reduced to $68.0\,\text{kJ}\times\text{mol}^{-1}$. Take $R = 8,314\,\text{J}\times\text{mol}^{-1}\times\text{K}^{-1}$ and a temperature of $500\,\text{K}$.

  1. a) Determine the activation energy of the reverse reaction without a catalyst and with a catalyst. State whether the presence of the catalyst changes the enthalpy of reaction or the equilibrium constant.
  2. b) In a simplified estimate of the energy barrier, the Boltzmann factor $\exp(-E_a/(R\times T))$ is used. Calculate the ratio between these factors for the catalyzed and uncatalyzed routes in the forward direction at $500\,\text{K}$. This factor is not the complete energy distribution, nor does it by itself guarantee an effective collision.
Solve

Back to the start: Iron powder burns in seconds because the fine grains expose a huge surface area to oxygen. More contact means more collisions per second, so more collisions with enough energy and the right orientation, and the rate skyrockets.

Summary

See all the exercises in this section