Rate, Collisions and Activation
An iron bar left outside rusts over months. The same iron, ground into a fine powder and held in a flame, can burn in seconds. The chemistry is the same. What changes is how fast it happens. Chemical kinetics is the study of how fast reactions go, and why.
Reaction rate
Drop a strip of magnesium into hydrochloric acid. The bubbling is fierce in the first seconds and barely noticeable a few minutes later. How can we define and measure a speed that keeps changing?
The real curve of $[\mathrm{A}]$ versus time is not a straight line. At the start there is plenty of reactant and the rate is high. As A is used up, the rate drops. The slope of the secant line gives the average rate over an interval. The slope of the tangent gives the instantaneous rate. The rate of disappearance is the opposite of the slope, so it comes out positive. To get the rate of reaction, divide by the coefficient of A (here it is 1).
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 $0.064\,\text{mol}\times\text{L}^{-1}$.
- a) Calculate the average rate of disappearance of $\mathrm{N_2O_5}$ over that time interval.
- b) Determine the overall average rate of the reaction normalized by the stoichiometry, and the average rate of formation of $\mathrm{NO_2}$.
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.
- a) Calculate the mean rate of disappearance of $\mathrm{A}$ over the three successive intervals: $[0, 10]\,\text{s}$, $[10, 20]\,\text{s}$ and $[20, 30]\,\text{s}$.
- b) Compare the three values numerically and explain, using collision theory, the cause of the variation observed as the reaction proceeds.
Collision theory
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 accomplish nothing?
Only a tiny fraction of collisions meet both requirements. That is why, even with trillions of collisions per second, most reactions are not instantaneous.
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})$$Use a simplified collision model with transfer of an oxygen atom between these molecules, without necessarily identifying it with the complete real mechanism:
- a) Explain why not all collisions between $\mathrm{CO}$ and $\mathrm{NO_2}$ molecules lead to the formation of products, stating the two essential requirements for a collision to be effective.
- b) Justify which of the two relative orientations favors the reaction: the carbon atom of $\mathrm{CO}$ striking an oxygen atom of $\mathrm{NO_2}$, or the oxygen atom of $\mathrm{CO}$ striking the nitrogen atom of $\mathrm{NO_2}$.
In the industrial synthesis of ammonia: $$\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:
$$\mathrm{N_2}\,(\mathrm{g}) + 3\,\mathrm{H_2}\,(\mathrm{g}) \rightleftharpoons 2\,\mathrm{NH_3}\,(\mathrm{g}) \quad \Delta H^\circ = -92.2\,\text{kJ}$$- a) Distinguish conceptually between the rate of a chemical reaction and its extent or equilibrium yield.
- 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.
Energy profiles and activation energy
The diagram shows an exothermic reaction: the products sit lower than the reactants, but the particles still have to climb the barrier $E_a$ first.
Look at the widget. Raising $E_a$ makes the hill taller (a slower reaction). Changing $\Delta H$ only moves the product level up or down relative to the reactants (exothermic or endothermic) and does not change the height of the peak.
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 $H_{\neq} = 130\,\text{kJ}\times\text{mol}^{-1}$.
- 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 ($E_{a,\text{inv}}$).
- 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.
Back to the start: Powdered iron burns in seconds because the fine grains expose far more surface to oxygen. That means many more collisions per second, so many more effective collisions that clear the activation energy.