Question

For a certain reaction ∆Hrxn = 54.2 kJ. Calculate the change in entropy for the surroundings (∆Ssurr) for the reaction at 25.0℃. (Assume constant pressure and temperature.)

Answer 1

The change in entropy for the surroundings (ΔSsurr) for a reaction that has an enthalpy change (ΔHrxn) of 54.2 kJ at 25.0°C is -182 J/K.

Step-by-step explanation:

To calculate the change in entropy for the surroundings (ΔSsurr) for a reaction with a given enthalpy change (ΔHrxn) of 54.2 kJ at 25.0°C, we use the formula ΔSsurr = -ΔHrxn / T. The temperature (T) must be in Kelvin, so we first convert 25.0°C to Kelvin (25 + 273.15 = 298.15 K). Substituting the values into the formula gives

ΔSsurr = - (54.2 kJ) / (298.15 K)

Converting kJ to J by multiplying 54.2 kJ by 1000, we get

ΔSsurr = - (54,200 J) / (298.15 K)

Calculating the entropy change yields

ΔSsurr = -182 J/K

The negative sign indicates that the entropy of the surroundings decreases when the reaction absorbs heat from the surroundings (endothermic process).