Exercises: Rate, collisions and activation

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 $\mathrm{NO_2}$.

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 $\mathrm{NO_2}$.

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.

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.

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.

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.