Question 84 of 92
Q.Calculate the entropy change during the melting of one mole of ice into water at 0°C and 1 atm pressure. Enthalpy of Fusion of ice is 6008 J mol^-1.
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2023Subjective· 2mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →For melting at constant temperature and pressure, Delta-S = Delta-H(fusion)/T; using Delta-H(fusion) = 6008 J/mol and T = 273 K gives Delta-S about 22.01 J K^-1 mol^-1.
At the melting point, ice and water coexist in equilibrium at constant temperature and pressure, so the process is reversible, and the entropy change of fusion is given by:
Delta-S(fusion) = Delta-H(fusion) / T
Given:
Delta-H(fusion) = 6008 J mol^-1
T = 0 degree C = 273 K (273.15 K, commonly rounded to 273 K)
Delta-S(fusion) = 6008 / 273 = 22.007 J K^-1 mol^-1, about 22.01 J K^-1 mol^-1
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