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18 grams of ice is converted into water at 0°C and 1 atm. The entropy of Hâ‚‚ and Hâ‚‚O is 38.2 J/mol·K and 60 J/mol·K, respectively. The enthalpy for this conversion is:

A. 595.14 J/mol
B. 595.14 J/mol
C. -595.14 J/mol
D. None of these

Answer :

Final answer:

The enthalpy change for the conversion of ice to water at 0°C and 1 atm can be calculated using the equation ΔH = ΔS. Given the provided entropies of Hâ‚‚ and Hâ‚‚O, we can calculate the entropy change. However, without information about the enthalpy of Hâ‚‚ and Hâ‚‚O, we cannot determine the exact enthalpy change.

Therefore the correct answer is D). None of these

Explanation:

The enthalpy change for the conversion of ice to water at 0°C and 1 atm can be calculated using the equation:

ΔH = ΔS

Where ΔH is the enthalpy change and ΔS is the entropy change. Given that the entropy of Hâ‚‚ is 38.2 J/mol K and the entropy of Hâ‚‚O is 60 J/mol K, we can calculate the entropy change of the conversion as follows:

ΔS = ΔSproducts - ΔSreactants = 60 J/mol K - 38.2 J/mol K = 21.8 J/mol K

Since the question asks for the enthalpy change, not the entropy change, and does not provide information about the enthalpy of Hâ‚‚ or Hâ‚‚O, we cannot calculate the exact enthalpy change using the given information. Therefore, the correct answer is D. None of these.

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Rewritten by : Jeany