In a drive toward cleaner energy production, fuel cell holds great promise because of its capability — Physical Chemistry Chemistry Question
Water-gas-shift Reaction
In a drive toward cleaner energy production, fuel cell holds great promise because of its capability of generating electricity directly from chemical reactions with environmentally-benign byproducts. In particular, for hydrogen fuel cell, the only waste produced by the device is just water.
In order to use fuel cell at an industrial scale, continuous production of hydrogen that directly feeds into a fuel cell module is required. One option to mass produce hydrogen for this purpose is the conversion of hydrocarbon fuel using hot steam. However, this kind of reaction often leads to mixed products that consist of H2, CO2, and CO. Moreover, CO is not only hazardous to human health, but it also degrades fuel cell’s active material. The reversible water-gas-shift (WGS) reaction, CO + H2O ⇌ CO2 + H2, provides one method of converting toxic CO into CO2 and useful H2. The efficiency of this reaction strongly depends on the solid catalyst used.
In one process, an equimolar mixture of CO and steam is continuously passed to a WGS reactor containing catalyst at atmospheric pressure and 0 °C. Assuming that the catalyst is 95.0% efficient in converting the reactants into the products and that the reaction in the reactor is approximately at equilibrium in this condition, estimate the free energy change for this reaction?
Model Answer
5.1 The mole fraction of H2 in the reactor, xH2 = 0.475. Thus, pH2 = 0.475 atm. And likewise, pCO2 = 0.475 atm and pH2O = pCO = 0.025 atm. Therefore, K = (0.475 × 0.475) / (0.025 × 0.025) = 3.6 × 10^2. Go = – RT ln K = – (8.314 J K-1 mol-1 × 273 K × ln(3.6×10^2)) = –13.4 kJ mol-1.
Now assume that a large surface of catalyst is initially available to reacting molecules, and the rate of reaction is measured immediately at the onset of the reaction. Below are the initial rates measured at different initial pressures of CO and H2O.
What is X?
Model Answer
5.2 The kinetics data given reflects the forward rate of the WGS reaction. The only rate law that is consistent with the given data is rf = kf pCO pH2O, and kf = 4.4 × 10-3 atm-1 s-1.
Thus, X = (4.4 × 10-3 atm-1 s-1)(0.28 atm)(0.72 atm) = 8.9 × 10-4 atm s-1.
In another condition where the pressure of hydrogen is found to be 0.50 atm, – dp(H2)/dt = 3.0×10-7 atm s-1. According to the information given in questions 5.1, 5.2, and this question, estimate the rate of hydrogen production when the pressures in the reactor of CO, H2O, CO2, and H2 are 0.14, 0.14, 0.36, and 0.36 atm, respectively. (Give your answer to three significant figures.)
Model Answer
5.3 kb = kf / K = (4.4 × 10-3 atm-1 s-1) / (3.6 × 10^2) = 1.2 × 10-5 atm-1 s-1.
So during the normal course of the reaction, r = rf - rb = kf pCO pH2O - kb pCO2 pH2 =
(4.4×10-3 atm-1 s-1)(0.14 atm)(0.14 atm) – (1.2×10-5 atm-1 s-1)(0.36 atm)(0.36 atm) =
= 8.44 × 10-5 atm s-1.
Calculate the Gibbs free energy change for the conditions described in question 5.3.
Model Answer
5.4 G = Go + RT ln Q = (-13.4 kJ mol-1) + (8.314 J mol-1 K-1)(273 K) ln [ (0.36)(0.36) / (0.14)(0.14) ] =
– 9.1 kJ mol-1
Surface coverage, θ, is an important kinetics parameter, especially for reactions on solid surfaces. It can be defined as the number of adsorbed molecules on a surface divided by the total number of adsorbing sites on that surface. For WGS, after the adsorption of CO and H2O on the catalyst’s surface, a carbonyl intermediate can be formed, which then dissociates to give surface-bound CO2 and H atom. If the carbon dioxide is produced at a rate of 1.0×1011 molecules s-1 cm-2 with associated rate constant of 2.0×1012 molecules s-1 cm-2, what is the value of θ for this intermediate?
Model Answer
5.5 The reaction CO2H(ads) → CO2(ads) + H(ads) is first-order, whose rate can be expressed as rate = k[CO2H] = k S0 = k′ , where S0 denotes the maximum number of adsorbed intermediates for this surface.
Thus, θ = rate/k′ = (1.0 × 1011 molecules s-1 cm-2) / (2.0 × 1012 molecules s-1 cm-2) = 0.050.