Question

Calculate the entropy change in the surroundings associated with this reaction occurring at 25∘C. Express the entropy change to three significant figures and include the appropriate units.

Answer 1

To calculate the entropy change of the surroundings for the reaction at 25°C, convert the enthalpy change to J/mol and use the equation ΔS = -ΔH/T. The result is an entropy change of 630 J/(mol·K) for the surroundings.

Step-by-step explanation:

To calculate the entropy change (ΔS) in the surroundings for a chemical reaction at a constant temperature, we use the equation ΔS = -ΔH/T, where ΔH is the enthalpy change and T is the temperature in Kelvin. Given the enthalpy change (ΔH) of the reaction H2(g) + O2(g) = H2O2(l) is -187.78 kJ/mol, and the reaction is at 25°C (which is 298 K), the entropy change in the surroundings can be calculated.

Firstly, convert ΔH to the same units as entropy, which is J/mol: -187.78 kJ/mol * 1000 J/kJ = -187780 J/mol. Then, insert the values into the equation to find ΔS: ΔS = - (-187780 J/mol) / 298 K = 630.4 J/(mol·K). To express this to three significant figures, ΔS = 630 J/(mol·K), indicating an increase in the entropy of the surroundings since the value is positive.

Answer 2

That means that if you are calculating entropy change, you must multiply the enthalpy change value by 1000. So if, say, you have an enthalpy change of -92.2 kJ mol-1, the value you must put into the equation is -92200 J mol-1