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 the calcite form of CaCO3 to the aragonite form is +0.21 kJ. The enthalpy change (in J) in the conversion at 2.7 bar is (The densities of the solids calcite and aragonite are 2.7 g cm-3 and 3.0 g cm-3, respectively.)
Answer:
209.00
Solution:
M(CaCO3) = 100 g/mol. Δ V = 100(1/3.0-1/2.7) = 33.33 - 37.04 = -3.70 cm3 = -3.70×10-6 m3. PΔ V = 2.7×105×(-3.70×10-6) = -1.0 J. Δ H = 210 - 1.0 = 209 J. Ans = 0209.
Stream:JEESubject:ChemistryTopic:Thermodynamics - ISubtopic:DH/DU for reaction
⏱ 2mℹ️ Source: QB
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