Chemistry - Thermodynamics - I Question with Solution | TestHub
ChemistryThermodynamics - IDH/DU for reactionHard2 minQB
ChemistryHardnumerical
The internal energy change in the conversion of 1 mole of rhombic sulfur (S, 32 g/mol) to monoclinic sulfur is +0.350 kJ. Find the enthalpy change (in J, nearest integer) at a constant pressure of 100 bar. The densities of the solids rhombic and monoclinic sulfur are 2.07 g cm-3 and 1.96 g cm-3, respectively.
Answer:
359.00
Solution:
Δ V = 32(1/1.96-1/2.07) = 32(0.51020-0.48309) = +0.868 cm3 = 8.68×10-7 m3. PΔ V = 100×105×8.68×10-7 = 8.68 J (1 bar = 105 Pa, not 101.3 kPa). Δ H = 350 + 8.68 = 358.7 ≈ 359 J. Ans = 0359.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:DH/DU for reaction
⏱ 2mℹ️ Source: QB
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