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Chemical transformations · Rate and mechanisms

Factors, Catalysts, and Experimental Design

An effervescent tablet fizzes away in seconds, but the same tablet sitting in one solid lump would take much longer. The chemistry is identical; only the way the tablet meets the water has changed. How can you speed up a reaction, slow it down, or test what controls it? That is what this page is about.

Factors that affect reaction rate

A campfire burns slowly with thick logs but flares up with wood shavings. A cut apple browns in minutes on the counter but lasts days in the refrigerator. What is happening at the particle level to change the pace so much?

Remember collision theory: for a reaction to happen, particles must collide with enough energy (at least the activation energy, $E_a$) and with the right orientation. Anything that increases the number of effective collisions per second speeds up the reaction.

A single solid block and the same block broken into four pieces of equal total mass, showing that more surface area is exposed after crushing one block crush less surface 4 pieces, same mass more exposed area → faster reaction
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Consider the heterogeneous reaction between solid calcium carbonate and an aqueous solution of hydrochloric acid:…

Consider the heterogeneous reaction between solid calcium carbonate and an aqueous solution of hydrochloric acid: $\mathrm{CaCO_3(s) + 2\,HCl(aq) \rightarrow CaCl_2(aq) + CO_2(g) + H_2O(l)}$. To study its kinetics at $25\,^\circ\mathrm{C}$, two independent experiments are carried out, using in both exactly $5.0\,\mathrm{g}$ of $\mathrm{CaCO_3}$ and $100\,\mathrm{mL}$ of aqueous solution of $\mathrm{HCl}\ 1.0\,\mathrm{mol}\times\mathrm{L}^{-1}$. In the first experiment a single compact piece of marble is used and in the second, finely powdered marble.

  1. a) Justify, using collision theory, in which of the two experiments the initial rate of gas evolution will be greater.
  2. b) Explain, with reasoning, whether the total volume of $\mathrm{CO_2(g)}$ released at the end of the reaction will differ in the two cases.
Solve

Catalysts: same destination, different path

Turn the catalyst on. The peak of the energy profile drops, but the starting level (reactants) and ending level (products) do not move. So $\Delta H$ is unchanged.

A catalyst lowers the barrier for the forward reaction and for the reverse reaction by the same amount. That is why a reversible system reaches equilibrium sooner, but arrives at the same equilibrium.

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In the industrial synthesis of sulfur trioxide by the contact process, sulfur dioxide is oxidized at $450\,^\circ\mathrm{C}$ through an exothermic equilibrium:…

In the industrial synthesis of sulfur trioxide by the contact process, sulfur dioxide is oxidized at $450\,^\circ\mathrm{C}$ through an exothermic equilibrium: $\mathrm{2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g)}$, with $\Delta H^\circ = -198\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. To increase the rate of the process, solid divanadium pentoxide, $\mathrm{V_2O_5(s)}$, is used as a catalyst.

  1. a) Explain how $\mathrm{V_2O_5(s)}$ modifies the activation energy of the forward reaction and of the reverse reaction.
  2. b) Explain whether the introduction of the catalyst increases the standard enthalpy of the reaction $\Delta H^\circ$ or the concentration of $\mathrm{SO_3(g)}$ reached at chemical equilibrium at that temperature.
Solve
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The attached figure compares the potential energy profiles against the reaction coordinate for the same gas-phase transformation carried out by an uncatalyzed pathway…

The attached figure compares the potential energy profiles against the reaction coordinate for the same gas-phase transformation carried out by an uncatalyzed pathway and by a pathway with a homogeneous catalyst:

120 100 80 60 40 20 0 Potential energy (kJ/mol) Reaction coordinate Reactants Products Uncatalyzed pathway Catalyzed pathway

. In this graph, the enthalpy level of the reactants is at $20\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ and that of the products at $0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. The peak of the uncatalyzed profile reaches $100\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, whereas that of the catalyzed process lies at $60\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.

  1. a) Determine analytically the activation energies of the forward and reverse reactions for the uncatalyzed pathway and for the catalyzed pathway.
  2. b) Calculate the enthalpy change of the reaction ($\Delta H$) and explain whether the catalyst changes the position of the chemical equilibrium.
Solve
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The decomposition of hydrogen peroxide, $\mathrm{2\,H_2O_2(aq) \rightarrow 2\,H_2O(l) + O_2(g)}$, can be accelerated by the enzyme catalase (a biocatalyst) or by solid…

The decomposition of hydrogen peroxide, $\mathrm{2\,H_2O_2(aq) \rightarrow 2\,H_2O(l) + O_2(g)}$, can be accelerated by the enzyme catalase (a biocatalyst) or by solid manganese dioxide, $\mathrm{MnO_2(s)}$ (an inorganic catalyst). In a series of trials carried out between $10\,^\circ\mathrm{C}$ and $70\,^\circ\mathrm{C}$ it is observed that, with $\mathrm{MnO_2}$, the reaction rate increases continuously as the temperature rises. However, with catalase, the rate reaches a maximum value at around $37\,^\circ\mathrm{C}$ and falls sharply until it disappears completely at $65\,^\circ\mathrm{C}$.

  1. a) Explain at the molecular level why the enzyme loses its activity at high temperatures, in contrast to the inorganic catalyst.
  2. b) Define the concept of enzyme specificity and state one fundamental advantage of using enzymes in chemical and industrial synthesis.
Solve

Designing a kinetics experiment

To find out how one factor affects rate, you need a controlled experiment.

Diagram of the independent variable, the dependent variable, and the control variables in a kinetics experiment independent [HCl] changes dependent gas volume / time controls T, mass, grain size, volume
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A school technical report on the Haber-Bosch process for the synthesis of ammonia, $\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}$ (with…

A school technical report on the Haber-Bosch process for the synthesis of ammonia, $\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}$ (with $\Delta H^\circ = -92.2\,\mathrm{kJ}\times\mathrm{mol}^{-1}$ at $298\,\mathrm{K}$), states: "To maximize the amount of ammonia produced in the reactor it is advisable to operate at very high temperatures (above $900\,^\circ\mathrm{C}$), since raising the temperature makes the reaction faster and therefore the percentage of ammonia obtained at equilibrium will be higher. Likewise, adding an iron catalyst with promoters shifts the equilibrium towards the products, increasing the final yield of ammonia."

  1. a) Identify and refute, with physico-chemical rigor, the conceptual errors the report contains about the effect of temperature on equilibrium.
  2. b) Correct the statement about the role of the catalyst, explaining the compromise between kinetics and thermodynamics required in the chemical industry.
Solve
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You are to design a safe laboratory experiment to determine quantitatively how the concentration of a reactant affects the initial rate of the reaction:…

You are to design a safe laboratory experiment to determine quantitatively how the concentration of a reactant affects the initial rate of the reaction: $\mathrm{CaCO_3(s) + 2\,HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}$. The available equipment is marble in homogeneous chips, aqueous solution of $\mathrm{HCl}\ 1.0\,\mathrm{mol}\times\mathrm{L}^{-1}$, distilled water, $100\,\mathrm{mL}$ Erlenmeyer flasks, a $50\,\mathrm{mL}$ gas syringe with an airtight connection, a stopwatch, graduated cylinders, a precision balance and a thermostatic water bath at $25\,^\circ\mathrm{C}$.

  1. a) Describe the experimental design with reasoning, identifying precisely the independent variable, the dependent variable and at least three essential control variables that must be kept strictly constant.
  2. b) Explain the analytical procedure for calculating the initial rate and state the relevant safety precautions for this experiment. The design is to be a proposal for a supervised school lab: size the quantities so as not to exceed the capacity of the syringe and avoid blocking its movement.
Solve
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Topic challenge. How would you optimize the speed of action of an effervescent medicine? Design and carry out a home kinetic study with effervescent tablets (or baking…

Topic challenge. How would you optimize the speed of action of an effervescent medicine? Design and carry out a home kinetic study with effervescent tablets (or baking soda) and water:

  1. Experimental phase: Use your phone's stopwatch to measure the time ($t$) for complete reaction in four trials: a whole tablet in cold water (check the temperature of your refrigerator or a thermometer, $\approx 5.0\text{ }^\circ\text{C}$), whole at room temperature (record the one from the National Weather Service for your town or an indoor thermometer, $\approx 20.0\text{ }^\circ\text{C}$), crushed at room temperature, and whole with a few drops of vinegar added.

  2. Data processing: Construct a data table with your measurements and calculate the relative mean rate ($v_m \propto 1/t$). Justify the changes observed according to collision theory and the activation energy ($E_a$).

  3. Modeling and mechanism: Propose a rate law ($v = k[\text{A}]^\alpha$), consistent with your mechanism, justifying the order of reaction with respect to the reactants (your trials do not vary concentrations, so they cannot determine it experimentally), and formulate a hypothetical two-step mechanism in which you identify the rate-determining step and discuss the role of the water and the acid (solvent, reactant or catalyst).

Submit a brief one-page report with your tables, calculations and conclusions applied to the pharmaceutical field.

Assessment criteria:

  • Rigor in the home experimental design and precision in recording your own data.

  • Coherent explanation of the kinetic factors using collision theory and $E_a$.

  • Appropriate physico-chemical justification of the rate law and of the proposed step-by-step mechanism.

Solve

Beyond the standards: the Arrhenius equation

Chemists describe how the rate constant $k$ depends on temperature with the Arrhenius equation:

$$k = A\,e^{-E_a/(RT)}$$

Here $A$ is the frequency factor (related to how often collisions happen with the right orientation), $R = 8.314\ \mathrm{J/(mol\cdot K)}$, and $T$ is in kelvins.

Raise $E_a$ with the slider: the curve stays near zero until much higher temperatures, then climbs steeply. Because $E_a$ sits in an exponent divided by $T$, a small temperature rise can change $k$ a lot. That is why many reactions roughly double or triple in rate for each $10\,^\circ\mathrm{C}$ increase.

Summary

Back to the start: the effervescent tablet dissolves faster crushed because crushing increases the surface area, so more collisions happen each second.

See all the exercises in this section