Chemistry - Thermodynamics - II Question with Solution | TestHub
Match the processes in Column I with the properties in Column II. Column I (process): (A) Raw egg to boiled egg; (B) CaCO3(s) to CaO(s) + CO2(g); (C) White phosphorus(s) to red phosphorus(s); (D) H2(g) adsorption on Pd. Column II (property): (p) Positive enthalpy change; (q) Positive entropy change; (r) Negative Gibbs-energy change; (s) Allotropic change; (t) Negative entropy change. Conditions: Cook the egg near 100 C at 1 atm; consider its net endothermic denaturation. For CaCO3, heat above the equilibrium decomposition temperature at the imposed CO2 pressure. Compare white and red phosphorus near 298 K under standard conditions, without air. For H2, consider spontaneous adsorption on clean Pd before adsorption equilibrium.
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Solution:
(A) At cooking temperature, the specified net denaturation absorbs heat, increases entropy, and proceeds spontaneously: p,q,r. This is not a claim that raw eggs cook spontaneously at room temperature. (B) CaCO3 decomposition absorbs heat and produces gaseous CO2, increasing entropy. Above its decomposition threshold at the stated CO2 pressure, Gibbs energy decreases: p,q,r. (C) White to red phosphorus is an allotropic change. At standard conditions red phosphorus is thermodynamically more stable, with lower enthalpy and entropy than white phosphorus: r,s,t. Heating overcomes a kinetic barrier; it does not make the enthalpy change positive. (D) Spontaneous gas adsorption on clean Pd lowers Gibbs energy and removes gas-phase freedom, so entropy decreases: r,t. Adsorption is exothermic, so p does not apply. The complete mapping is A-p,q,r; B-p,q,r; C-r,s,t; D-r,t. Only revised option C matches.
