Factors affecting rate, catalysts and experimental design
A fizzy vitamin tablet vanishes in a glass of water in under a minute, but the same tablet sitting in your schoolbag stays solid for months. Crush it and it fizzes even faster. Nothing about the chemistry has changed, so what exactly is different, and how could you prove it in the lab?
Factors that affect the rate
A campfire burns slowly with big logs but flares up with thin kindling. A cut apple browns in minutes on the counter but lasts days in the fridge. The same reaction can run at very different speeds depending on how you set it up.
Collision theory explains why. For two particles to react they must collide with enough energy (at least the activation energy, $E_a$) and with the correct orientation. Anything that increases the number of effective collisions per second increases the rate of reaction.
Concentration and pressure
For gases, squeezing the mixture into a smaller volume at constant temperature raises the concentrations (halving $V$ doubles every concentration), so the rate rises.
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.
- a) Justify, using collision theory, in which of the two experiments the initial rate of gas evolution will be greater.
- 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.
The gas-phase synthesis of nitrogen dioxide takes place at $300\,^\circ\mathrm{C}$ according to the overall reaction:…
The gas-phase synthesis of nitrogen dioxide takes place at $300\,^\circ\mathrm{C}$ according to the overall reaction: $\mathrm{2\,NO(g) + O_2(g) \rightarrow 2\,NO_2(g)}$. In a closed reactor fitted with a movable piston at $300\,^\circ\mathrm{C}$, a mixture is introduced initially containing $0.20\,\mathrm{mol}$ of $\mathrm{NO}$ and $0.10\,\mathrm{mol}$ of $\mathrm{O_2}$, occupying an initial volume $V_1$.
- a) The experimental law is $v=k[\mathrm{NO}]^2[\mathrm{O_2}]$. State the partial orders and the overall order, and explain why these exponents require experimental evidence for an overall reaction.
- b) Deduce quantitatively by what factor the initial reaction rate is multiplied if the piston is moved so that the volume of the reactor is halved ($V_2 = V_1 / 2$), keeping the temperature of the system constant.
The Arrhenius equation: putting numbers on temperature
The rate constant for the gas-phase thermal decomposition of ethanal, $\mathrm{CH_3CHO(g) \rightarrow CH_4(g) + CO(g)}$, has been measured experimentally at two…
The rate constant for the gas-phase thermal decomposition of ethanal, $\mathrm{CH_3CHO(g) \rightarrow CH_4(g) + CO(g)}$, has been measured experimentally at two different temperatures, giving the values $k_1 = 1.10\times 10^{-2}\,\mathrm{L}\times\mathrm{mol}^{-1}\times\mathrm{s}^{-1}$ at $700\,\mathrm{K}$ and $k_2 = 7.95\times 10^{-1}\,\mathrm{L}\times\mathrm{mol}^{-1}\times\mathrm{s}^{-1}$ at $800\,\mathrm{K}$. Assume that the Arrhenius equation applies, with a constant pre-exponential factor $A$ over this temperature range. Data: universal gas constant $R = 8,314\,\mathrm{J}\times\mathrm{mol}^{-1}\times\mathrm{K}^{-1}$.
- a) Calculate the activation energy of the reaction in $\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- b) Determine the numerical value and the units of the pre-exponential factor $A$. You may use $k=A\exp(-E_a/(RT))$.
The thermal decomposition of a gaseous substance has an activation energy of $85.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. When the reaction is carried out in the presence…
The thermal decomposition of a gaseous substance has an activation energy of $85.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. When the reaction is carried out in the presence of a platinum mesh acting as a heterogeneous catalyst, the activation energy decreases to $52.0\,\mathrm{kJ}\times\mathrm{mol}^{-1}$. Assume that the pre-exponential factor $A$ in the Arrhenius equation remains identical in both processes and does not vary with temperature. Data: $R = 8,314\,\mathrm{J}\times\mathrm{mol}^{-1}\times\mathrm{K}^{-1}$.
- a) Calculate the numerical factor ($k_{\mathrm{cat}} / k_{\mathrm{no\,cat}}$) by which the rate constant increases when the catalyst is used, if the reactor is kept at $298\,\mathrm{K}$.
- b) Determine the absolute temperature to which the uncatalysed reaction would have to be heated for its rate constant to be exactly equal to that of the catalysed reaction at $298\,\mathrm{K}$. You may use $k=A\exp(-E_a/(RT))$.
Catalysts: same destination, different route
On an energy profile diagram, the reactants and products stay at the same levels, so $\Delta H$ is unchanged. Only the height of the peak drops. Toggle the catalyst in this widget:
This profile is for an exothermic reaction ($\Delta H < 0$). For an endothermic reaction the products lie above the reactants, as in the next widget. The catalyst lowers the peak in the same way, and $\Delta H$ is still unchanged.
Because the forward and reverse barriers both fall by the same amount, a catalyst speeds up both directions equally. So it does not change the position of equilibrium, the equilibrium yield or $\Delta H$. It only gets the system to equilibrium sooner.
How do catalysts work?
There are two general mechanisms:
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Surface adsorption (heterogeneous catalysis). The catalyst is a solid, usually a metal or metal oxide, and the reactants are gases or liquids. Reactant molecules adsorb onto the surface, their bonds weaken, and they are held close together in a good orientation. They react, and the products desorb, leaving the surface free again. Examples: $\mathrm{V_2O_5(s)}$ in the contact process, iron in the Haber process, platinum, palladium and rhodium in a car's catalytic converter.
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Formation of an intermediate (homogeneous catalysis). The catalyst is in the same phase as the reactants. It reacts with a reactant to form an intermediate, which then reacts further and regenerates the catalyst. Each step has a lower activation energy than the uncatalysed reaction.
Enzymes are biological catalysts (proteins). The reactant (the substrate) fits into the active site of the enzyme, forms an enzyme–substrate complex, reacts and is released as product. Enzymes are very specific. Their activity depends on temperature and pH: too hot and the protein is denatured, losing its shape and its activity. Catalase, found in liver and potato, speeds up the breakdown of hydrogen peroxide: $\mathrm{2\,H_2O_2(aq) \rightarrow 2\,H_2O(l) + O_2(g)}$.
In the industrial synthesis of sulfur trioxide by the contact process, sulfur dioxide is oxidised at $450\,^\circ\mathrm{C}$ through an exothermic equilibrium:…
In the industrial synthesis of sulfur trioxide by the contact process, sulfur dioxide is oxidised 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.
- a) Explain how $\mathrm{V_2O_5(s)}$ modifies the activation energy of the forward reaction and of the reverse reaction.
- 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.
The attached figure compares the potential energy profiles against the reaction coordinate for the same gas-phase transformation carried out by an uncatalysed pathway…
The attached figure compares the potential energy profiles against the reaction coordinate for the same gas-phase transformation carried out by an uncatalysed pathway and by a pathway with a homogeneous catalyst:
. 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 uncatalysed profile reaches $100\,\mathrm{kJ}\times\mathrm{mol}^{-1}$, whereas that of the catalysed process lies at $60\,\mathrm{kJ}\times\mathrm{mol}^{-1}$.
- a) Determine analytically the activation energies of the forward and reverse reactions for the uncatalysed pathway and for the catalysed pathway.
- b) Calculate the enthalpy change of the reaction ($\Delta H$) and explain whether the catalyst changes the position of the chemical equilibrium.
Designing kinetic experiments
To test how one factor affects the rate, you need a fair test.
Three standard investigations
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$\mathrm{CaCO_3 + HCl}$. Collect $\mathrm{CO_2}$ in a gas syringe (or measure the loss of mass on a balance) and time it. Plot volume against time. When you read a rate graph (volume or mass against time), the curve is steepest at the start and flattens as the reactant is used up. The rate at any moment is the slope of the tangent (the instantaneous rate); a flat curve means the reaction has stopped. The initial rate is the slope at $t = 0$. The average rate is $\Delta V/\Delta t$ over an interval.
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$\mathrm{Na_2S_2O_3 + HCl}$ (the "disappearing cross"). Sulfur forms and the solution turns cloudy: $\mathrm{Na_2S_2O_3(aq) + 2\,HCl(aq) \rightarrow 2\,NaCl(aq) + S(s) + SO_2(g) + H_2O(l)}$. Time how long it takes for a cross under the flask to vanish. Since the same amount of sulfur is needed each time, the rate is proportional to $1/t$.
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Decomposition of $\mathrm{H_2O_2}$. Measure the volume of $\mathrm{O_2}$ over time with and without $\mathrm{MnO_2}$ (or catalase from liver or potato), and at different temperatures.
Good practice
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Safety and risk assessment: wear safety glasses, treat dilute acids with care, and use a fume cupboard if $\mathrm{SO_2}$ could be produced. Use small quantities of reactant so the gas produced does not push the plunger out of the syringe.
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Precision and accuracy: use a measuring cylinder, pipette or burette suited to the volume, a stopwatch, and a thermostatic water bath. Repeat and take a mean to reduce random errors. A thermometer that reads wrongly every time is a systematic error.
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Analyse the data: plot a graph, draw tangents carefully, and state your conclusion using the evidence only.
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 $65\,^\circ\mathrm{C}$.
- a) Explain at the molecular level why the enzyme loses its activity at high temperatures, in contrast to the inorganic catalyst.
- b) Define the concept of enzyme specificity and state one fundamental advantage of using enzymes in chemical and industrial synthesis.
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, measuring cylinders, a precision balance and a thermostatic bath at $25\,^\circ\mathrm{C}$.
- 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.
- 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 laboratory: size the quantities so as not to exceed the capacity of the syringe and avoid blocking its movement.
Topic challenge. How would you optimise the speed of action of an effervescent medicine? Design and carry out a home kinetic study with effervescent tablets (or…
Topic challenge. How would you optimise the speed of action of an effervescent medicine? Design and carry out a home kinetic study with effervescent tablets (or bicarbonate of soda) and water:
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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 fridge or a thermometer, $\approx 5.0\text{ }^\circ\text{C}$), whole at room temperature (record the one from Met Éireann for your locality or an indoor thermometer, $\approx 20.0\text{ }^\circ\text{C}$), crushed at room temperature, and whole with a few drops of vinegar added.
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Data processing: Construct a 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 ($E_a$).
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Modelling 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:
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Rigour in the home experimental design and precision in recording your own data.
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Coherent explanation of the kinetic factors using collision theory and $E_a$.
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Appropriate physico-chemical justification of the rate law and of the proposed step-by-step mechanism.
Back to the start: the fizzy tablet reacts fast in water because its acid and bicarbonate dissolve and can meet as ions in solution; crushing it increases the surface area further, and warm water adds more energy. In your schoolbag the acid and bicarbonate are dry and locked in the solid, so they can hardly collide and react extremely slowly.