Latent heart of fusion of ice is 0.333kJ/g. The increase in entropy when 1mole water melts at 0°C will be?
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Data:
Latent heat of fusion of ice = 0.333 k J/g = 333 J/g
Amount of ice = 1 mole
Mass of 1 mole of ice will be the same as mass of 1 mole of water.
So,
Mass of 1 mole of ice = 2 (for 2 moles of H₂ atoms) + 16 (for 1 mole of O atom)
Mass of 1 mole of ice = 18 g
Temperature = T = 0°C = 273.15 K
Change in entropy = ?
Solution:
Entropy change = ΔS = n ΔH / T
Entropy change = ΔS = (18) ( 333) / 273.15
Entropy change = ΔS = 21.94 J/K
which is the required answer.
Hopefully it helps.
Thanks.
Latent heat of fusion of ice = 0.333 k J/g = 333 J/g
Amount of ice = 1 mole
Mass of 1 mole of ice will be the same as mass of 1 mole of water.
So,
Mass of 1 mole of ice = 2 (for 2 moles of H₂ atoms) + 16 (for 1 mole of O atom)
Mass of 1 mole of ice = 18 g
Temperature = T = 0°C = 273.15 K
Change in entropy = ?
Solution:
Entropy change = ΔS = n ΔH / T
Entropy change = ΔS = (18) ( 333) / 273.15
Entropy change = ΔS = 21.94 J/K
which is the required answer.
Hopefully it helps.
Thanks.
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