The heat of vaporization AHv, of ammonia (NH3) is 23.4 kJ/mol. Calculate the change in entropy AS when 23. g of ammonia boils at -33.5 °C.
Be sure your answer contains a unit symbol and the correct number of significant digits.

Respuesta :

Answer:

ΔS(23g NH₃) = 132.2 j/K  ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).

Explanation:

Given ΔHv(NH₃) = 23.4 Kj/mol = 23,400 joules/mole

T = -33.5°C = (-33.5 + 273)K = 239.5K

molar ΔS (at boiling) = ΔHv/T = 23,400 j/mol / 239.5K = 97.70 j/mol·K

ΔS(23g NH₃) = molar ΔS (at boiling) x (moles NH₃) = 97.70j/mol·K  x  (23g/17g·mol⁻¹) = 132.2 j/K ≅ 130 j/K (~2 sig. figs. based on 2 sig. figs. in 23 grams NH₃).

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