Ask Question
22 January, 21:14

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

+3
Answers (1)
  1. 23 January, 01:13
    0
    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “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 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers