[12%] Hydrogen gas reacts with oxygen gas to give water vapor with ∆H°rxn = –241.8 kJ mol-1 and ∆S°r — Thermodynamics Chemistry Question
Problem Context
[12%] Hydrogen gas reacts with oxygen gas to give water vapor with ∆H°rxn = –241.8 kJ mol-1 and ∆S°rxn = –44.5 J mol-1 K-1.
H2(g) + 0.5 O2(g) → H2O(g)
Calculate S° of H2O(g).
Model Answer
∆S°rxn = S° (H2O(g)) – S° (H2(g)) – 0.5•S° (O2(g))
–44.5 J mol-1 K-1 = S° (H2O(g)) – (130.7 J mol-1 K-1) – 0.5•(205.2 J mol-1 K-1)
S° (H2O(g)) = 188.8 J mol-1 K-1
The bond dissociation enthalpy (BDE) of an average O–H bond in water is 463 kJ mol-1 and the BDE of the H– H bond in H2 is 436 kJ mol-1. What is the BDE of the O=O bond in O2?
Model Answer
∆H°rxn = 0.5 × BDE(O=O) + BDE(H–H) – 2 × BDE(O–H)
–242 kJ mol-1 = 0.5 × BDE(O=O) + 436 kJ mol-1 – 2 × (463 kJ mol-1)
BDE(O=O) = 496 kJ mol-1
0.100 mol H2(g) and 0.100 mol O2(g), both initially at 100 °C, react completely in a sealed vessel that is maintained at 1 bar pressure. The vessel is machined as part of a 1.00-kg block of aluminum (Cp = 0.89 J g-1 K-1), which efficiently absorbs the heat generated in the reaction but which is well insulated from its surroundings. What is the final temperature of the aluminum block and the contents of the reaction vessel? Assume that Cp and ΔH°rxn are independent of temperature.
Model Answer
Reacting 0.1 mol H2 (the limiting reagent) will produce 24.18 kJ of heat (at constant pressure). The contents of the vessel after reaction are 0.1 mol H2O and 0.05 mol O2, so the total heat capacity of the reaction contents plus the Al block is (1.802 g H2O) × (4.18 J g-1 K-1) + (1.60 g O2) × (0.92 J g-1 K-1) + (1000 g) × (0.89 J g-1 K-1) = 899 J K-1.
∆T = q/Cp = (24180 J) / (899.0 J K-1) = 26.9 K
Since the initial temperature is 373 K, the final temperature is 400 K (127 °C).
Humid air at 298 K has an overall pressure of 1.0 bar and is 20. vol% O2 and 3.1 vol% H2O. What is the minimum volume percentage of H2(g) in humid air necessary for its combustion to be spontaneous?
Model Answer
∆G = ∆G° + RTln(Q). At 298 K, ∆G° = ∆H° – T∆S° = –241.8 kJ mol-1 – (298 K)(–0.0445 kJ mol-1 K-1) = –228.5 kJ mol-1.
For the reaction to just be spontaneous, ∆G = 0, so Q = e^(–∆G°/RT) = 1.1 × 10^40.
Q = 1.1 × 10^40 = p(H2O) / (p(H2)•p(O2)^(1/2))
Substituting in p(H2O) = 0.031 bar and p(O2) = 0.20 bar gives
p(H2) = 6.1 × 10^-42 bar
This corresponds to 6.1 × 10^-40 volume% of H2. (This is much, much less than one molecule per liter, so any amount of hydrogen in humid air would react spontaneously to form water vapor.)
The lower flammability limit (LFL) of a substance is its minimum volumetric concentration needed to propagate a flame under a given set of conditions. The LFL of hydrogen gas in humid air is approximately 4%. Account for the difference between your answer in part d and the experimental value for the LFL.
Model Answer
The thermodynamic spontaneity of the reaction need not correspond to the kinetic conditions required to support a self-sustaining flame. Evidently at concentrations below about 4 vol%, the reaction cannot proceed rapidly enough to produce enough reactive intermediates to support continued rapid reaction.