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What is the change in entropy of the surroundings (ΔSsurr) for the reaction of 2 SO2(g) + O2(g) → 2 SO3(g) at 25.0 °C, given that ΔH = -196 kJ?
A
-196 J/K
B
656 J/K
C
-656 J/K
D
196 J/K
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1
Understand that the change in entropy of the surroundings (ΔSsurr) is related to the heat exchanged with the surroundings and the temperature at which the process occurs. The formula to use is ΔSsurr = -ΔH/T.
Convert the given enthalpy change (ΔH) from kilojoules to joules. Since 1 kJ = 1000 J, multiply -196 kJ by 1000 to get -196,000 J.
Identify the temperature at which the reaction occurs. The problem states that the temperature is 25.0 °C. Convert this temperature to Kelvin by adding 273.15 to the Celsius temperature: 25.0 + 273.15 = 298.15 K.
Substitute the values into the formula ΔSsurr = -ΔH/T. Use ΔH = -196,000 J and T = 298.15 K.
Calculate the change in entropy of the surroundings (ΔSsurr) using the formula ΔSsurr = -ΔH/T. This will give you the value in J/K.