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 mean rate of disappearance of $\mathrm{N_2O_5}$ over that time interval.
2. b) Determine the overall mean rate of the reaction normalised by the stoichiometry, and the mean rate of formation of $\mathrm{NO_2}$.

The decomposition of reactant $\mathrm{A}$ in aqueous solution according to $\mathrm{A}\,(\mathrm{aq}) \rightarrow \mathrm{B}\,(\mathrm{aq}) + \mathrm{C}\,(\mathrm{aq})$ 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.

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 favours 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 $\mathrm{NO_2}$.

Two reactors are designed to study two competing bimolecular pathways starting from the same gaseous reactant $\mathrm{X}$:

Pathway 1: $2\,\mathrm{X}\,(\mathrm{g}) \rightarrow \mathrm{Y}\,(\mathrm{g})$, exothermic with $\Delta H_1^\circ = -80\,\text{kJ}\times\text{mol}^{-1}$ and $E_{a,1} = 45\,\text{kJ}\times\text{mol}^{-1}$.

Pathway 2: $2\,\mathrm{X}\,(\mathrm{g}) \rightarrow \mathrm{Z}\,(\mathrm{g})$, exothermic with $\Delta H_2^\circ = -150\,\text{kJ}\times\text{mol}^{-1}$ and $E_{a,2} = 90\,\text{kJ}\times\text{mol}^{-1}$.

The steric factor for Pathway 1 is very restrictive ($p_1 = 0.01$) owing to the requirement for precise spatial alignment, whereas for Pathway 2 it is not very restrictive ($p_2 = 0.80$). The total collision frequency $Z$ is identical for both pathways.

1. a) Write the modified Arrhenius rate constant ($k = p\times Z\times\exp(-E_a/(R\times T))$) and determine the temperature at which both pathways would have exactly the same rate constant ($k_1 = k_2$).
2. b) Assuming the entropies of both products are equal, compare their thermodynamic stability using the Gibbs energy. Distinguish which pathway forms product faster initially at temperatures below or above the calculated threshold, and why this kinetic preference does not determine the final composition if both pathways are reversible and reach equilibrium. Data: $R=8,314\,\mathrm{J}\times\mathrm{mol}^{-1}\times\mathrm{K}^{-1}$.

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 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.

A gas-phase reaction proceeds along the uncatalysed 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 catalysed and uncatalysed 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.

To quantify the rate of oxidation of the iodide ion by hydrogen peroxide in acidic medium:

$$\mathrm{H_2O_2}\,(\mathrm{aq}) + 2\,\mathrm{I^-}\,(\mathrm{aq}) + 2\,\mathrm{H^+}\,(\mathrm{aq}) \rightarrow \mathrm{I_2}\,(\mathrm{aq}) + 2\,\mathrm{H_2O}\,(\mathrm{l})$$

three independent replicates are carried out in a buffered solution at $298\,\text{K}$. The measured concentrations of $\mathrm{I_2}$ formed after $120\,\text{s}$ are: Replicate 1: $(3.52 \pm 0.04)\times 10^{-3}\,\text{mol}\times\text{L}^{-1}$; Replicate 2: $(3.60 \pm 0.04)\times 10^{-3}\,\text{mol}\times\text{L}^{-1}$; Replicate 3: $(3.56 \pm 0.04)\times 10^{-3}\,\text{mol}\times\text{L}^{-1}$, starting in all cases from a zero initial concentration of $\mathrm{I_2}$. The time is measured with a stopwatch of uncertainty $\pm 1\,\text{s}$.

1. a) Determine the mean concentration of $\mathrm{I_2}$ at $t = 120\,\text{s}$ with a conservative estimate of the uncertainty, and calculate the mean rate of formation of $\mathrm{I_2}$ over that interval with its corresponding uncertainty. Use, as the convention for this exercise, $\Delta c=(c_{\max}-c_{\min})/2+0.04\times10^{-3}\,\mathrm{mol}\times\mathrm{L}^{-1}$ and $\Delta v/v=\Delta c/\bar c+\Delta t/t$.
2. b) Deduce the normalised mean rate of the reaction and the mean rate of consumption of the iodide ion $\mathrm{I^-}$ over that time interval, giving the correct units.