⚗️ Calculating Entropy Changes in the Surroundings

✔ https://StudyForce.com ✔ https://Biology-Forums.com ✔ Ask questions here: https://Biology-Forums.com/index.php?... Follow us: ▶ Facebook:   / studyforceps   ▶ Instagram:   / studyforceonline   ▶ Twitter:   / studyforceps   Q1. Consider the combustion of propane gas. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) ΔHrxn = −2044 kJ (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 °C. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Is the reaction spontaneous? What you'll need: Entropy of Surrounding: (ΔS_surr=(−ΔH_rxn)/T) Entropy of Universe: (ΔS_univ=ΔS_sys+ΔS_surr ) Q2. A reaction has ΔHrxn = −107 kJ and ΔSrxn = 285 J ∕ K. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings? Extra: Consider the reaction between nitrogen and oxygen gas to form dinitrogen monoxide. 2 N2(g) + O2(g) → 2 N2O(g) ΔHrxn = +163.2 kJ (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 °C. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Is the reaction spontaneous? Solution (a) The entropy change of the surroundings is given by Equation 17.3. Substitute the value of ΔHrxn and the temperature in kelvins and calculate ΔSsurr. (b) Determine the number of moles of gas on each side of the reaction. An increase in the number of moles of gas implies a positive ΔSsys. (c) The change in entropy of the universe is the sum of the entropy changes of the system and the surroundings. If the entropy changes of the system and surroundings are both the same sign, the entropy change for the universe also has the same sign.