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A reaction has a ΔHrxn=54.2 kJ. Calculate the change in entropy for the surroundings (ΔSsurr) for the reaction at 25.0 °C. (Assume constant pressure and temperature.)a. -0.184 J/K
b. -184 J/K
c. 1.66 J/K
d. 184 J/K

Respuesta :

Answer:

b. - 184 J/K

Explanation:

  • ΔSsurr = - ΔH/T....at constant P and T.

∴ ΔHrxn = 54.2 KJ = 54200 J

∴ T = 25°C ≅ 298.15 K

⇒ ΔSsurr = - (54200 J)/(298 K)

⇒ ΔSsurr = - 182 J/K ≅ - 184 J/K

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