The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

For the first part of the question, we have to identify which surface is the better heterogeneous catalyst. A good catalyst lowers the activation energy for a reaction, thus increasing the reaction rate. Since the activation energy without a catalyst is given as 335 kJ/mol, we need information about the activation energy with the catalysts on both surfaces. If not given, we cannot determine the better heterogeneous catalyst and explain why.

To find out how many times faster the reaction on the W surface is at 298 K compared to the reaction without a catalyst, we will use the Arrhenius equation to find the ratio of the rate constants. Remember that the Arrhenius equation is given by: \[k = A e^{-Ea / RT}\] Where k is the reaction rate constant, A is the frequency factor (assumed to be the same for both reactions), Ea is the activation energy, R is the ideal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin. In this case, we have both activation energies, no catalyst (Ea, no catalyst = 335 kJ/mol) and W surface (Ea, W = 100 kJ/mol), and the temperature T = 298 K. Now let's find the ratio of the rate constants without catalyst and on the W surface: \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{A e^{-Ea_{W} / RT}}{A e^{-Ea_{no\;catalyst} / RT}}\] By cancelling out A and simplifying, \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{e^{-Ea_{W} / RT}}{e^{-Ea_{no\;catalyst} / RT}}\] Now we can plug in the values of Ea and T to calculate the ratio: \[\frac{k_{W}}{k_{no\;catalyst}} = \frac{ e^{-\frac{100 \times 10^3}{8.314 \times 298}} }{ e^{-\frac{335 \times 10^3}{8.314 \times 298}} }\] After calculating the ratio: \[\frac{k_{W}}{k_{no\;catalyst}} \approx 7.4 \times 10^6\] This means the reaction on the W surface is approximately 7.4 million times faster than the reaction with no catalyst.

The given rate law for the decomposition reaction on both surfaces is: \[Rate = k \frac{[NH_3]}{[H_2]}\] To explain the inverse dependence of the rate on the H2 concentration, let's consider the mechanism of the reaction on these surfaces. A heterogeneous catalyst usually works by adsorbing the reactants onto its surface, weakening their bonds, and facilitating the reaction between them. In this case, NH3 and H2 are both adsorbed onto the surface of the catalyst, and the reaction between adsorbed NH3 and adsorbed H2 produces adsorbed N2 and H2. As the concentration of H2 increases, more H2 molecules get adsorbed onto the surface, occupying more sites and blocking NH3 from adsorbing onto those sites. As a result, there are fewer available sites for NH3 adsorption, which slows down the reaction between adsorbed NH3 and adsorbed H2. Hence, the rate of the reaction is inversely proportional to the concentration of H2.