for bismuth, By, the heat of vaporization at its normal boiling point of 1627 C

Question

for bismuth, By, the heat of vaporization at its normal boiling point of 1627 C is 172.0 kj/Mol. The entropy change when 2.38 moles of liquid Bi vaporizes at 1627 C, 1 ATM is for bismuth, By, the heat of vaporization at its normal boiling point of 1627 C is 172.0 kj/Mol. The entropy change when 2.38 moles of liquid Bi vaporizes at 1627 C, 1 ATM is for bismuth, By, the heat of vaporization at its normal boiling point of 1627 C is 172.0 kj/Mol. The entropy change when 2.38 moles of liquid Bi vaporizes at 1627 C, 1 ATM is

Explanation / Answer

Boiling point T = 1627 + 273 = 1900 K

Heat of vaporization Hvap = 172 kJ/mol

Moles of Bi = 2.38 mol

Hvap for 2.38 mol = (172 kJ/mol) x 2.38 mol = 409.36 kJ

Entropy change

S = Hvap/T

= 409.36 x 1000J /1900 K

= 215.45 J/K

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