One mole of ice is converted into water at 273 K and 1 atm. The entropies of (s) and (l) are 38.0 an — Thermodynamics and Thermochemistry Chemistry Question
Question
One mole of ice is converted into water at 273 K and 1 atm. The entropies of $H_2O$(s) and $H_2O$(l) are 38.0 and 58.0 J/K-mol, respectively. The enthalpy change for the conversion is
Answer: B
💡 Solution & Explanation
Since ice and water are in equilibrium at 273 K and 1 atm, Δ G = 0 => Δ H = T * Δ S. Δ S = S(liquid) - S(ice) = 58.0 - 38.0 = 20.0 J/K-mol. Δ H = 273 * 20 = 5460 J/mol.
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