Multiple Choice
A reaction has ΔrH = -107 kJ mol^-1 and ΔrS = 285 J K^-1 mol^-1. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?
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19. Chemical Thermodynamics
Entropy Calculations
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If the change in entropy of the surroundings for a process is -326 J/K at 439 K and constant pressure, what is the heat flow (ΔH) absorbed by the system?
A
-326 J
B
-143,314 J
C
143,314 J
D
326 J
Verified step by step guidance1
Understand that the change in entropy of the surroundings (ΔS_surroundings) is given as -326 J/K.
Recall the relationship between the change in entropy of the surroundings and the heat flow at constant pressure: ΔS_surroundings = -ΔH/T, where ΔH is the heat flow and T is the temperature in Kelvin.
Rearrange the formula to solve for ΔH: ΔH = -ΔS_surroundings * T.
Substitute the given values into the equation: ΔH = -(-326 J/K) * 439 K.
Calculate the product to find the heat flow (ΔH) absorbed by the system, ensuring the units are consistent.
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