4.8)  Calculate the values of ΔU, ΔH, and ΔS for the following process: 1 mole of liquid water at 25 C and 1 atm ➡️ 1 mole of steam at 100 C and 1 atm The molar heat of vaporization of water at 373 K is 40.79 kJ mol-1, and molar heat capacity of water is 75.3 J K-1 mol-1.  Assume molar heat capacity to be temperature independent and ideal-gas behavior.

Determine the universe entropy change ΔS° (universe) in J/K (with 4 significant figures) for the reaction below at 298.15 K. Note that the temperature of surroundings is also 298.15 K.  I2(s) + 3Cl2 (g) → 2 ICl3 (s) The standard entropy change and enthalpy change of the reaction (from the system) are given below:  ΔS° (reaction) = -450.33 J/K and ΔH° (reaction) = -179.16 kJ

When 5.93 g of potassium sulfate (K2SO4) is dissolved in 111 g of water in a styrofoam calorimeter of negligible heat capacity, the temperature drops from 25.00 to 23.21 °C. Based on this observation, calculate q for the water and AH° for the process, assuming that the heat absorbed by the salt is negligible. K2S04(s) 2K*(aq) + S04²- (aq) The specific heat of water is 4.184 J °C-1 g1. Give the answers in kJ. kJ AH20 ΔΗ kJ Not submitted

Consider the following reaction: 2 H2(g) + 2 NO(g) N2(g) + 2 H20(g) and the following table of standard enthalpies of formation and standard absolute entropies: compound AH (kJ mol-1) s° (J K-1 mol-1) H2(g) 130.8 NO(g) 91.3 210.8 N2(g) 191.6 H20(g) -241.8 188.8

Consider the reaction 2NaOH(aq) + H2(g)—2Na(s) + 2H2O(l) for which AH° = 368.6 kJ and AS° = 15.30 J/K at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.759 moles of NaOH(aq) react under standard conditions at 298.15 K. ASuniverse= J/K (2) Is this reaction reactant or product favored under standard conditions? (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'. Submit Answer

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Consider the reaction Fe(s) + 2HCl(aq) → FeCl₂ (s) + H₂(g) for which AH = -7.400 kJ and AS = 107.9 J/K at 298.15 K. 1. Calculate the entropy change of the UNIVERSE when 2.385 moles of Fe(s) react under standard conditions at 298.15 K. ASuniverse J/K 2. Is this reaction reactant or product favored under standard conditions? 3. If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored, choose 'reactant favored'. Submit Answer Retry Entire Group 5 more group attempts remaining

An insulated piston is compressed to 2 atms of pressure; it has 1 mol of a gas (Cv,m = 12.5 J/K/mol) and is at room temperature (25 °C). If you suddenly let go in a room at 1 atm of pressure, what are the final values of P, T, V, ΔU, ΔH, Δq, Δw, total ΔS, surroundings ΔS, and the system ΔS? Hint: remember that ΔU = Cv×(Tf - Ti) = -Pext x (Vf - Vi); what you don’t know is the finaltemperature and volume. You have to calculate these; note that the final volume is (nRTf) / (101.325 kPa)

The molar enthalpy of vaporization of ethanol is AvapH = 42.3 kJ/mol at its normal boiling point of 351.5 K. (a) Determine its molar entropy of vaporization at the boiling point, AvapS, in J/K. (b) The values of Cp for liquid and for gaseous ethanol for 300 < T< 400 are 112.4 and 70.85 J/mol-K, respectively. Use this information to estimate AvapH (in kJ/mol) at T = 320 K, assuming that the heat capacities of the liquid and gaseous states are constant over this temperature range. Hint: Construct a multi-step process that includes vaporization at 320 K, but includes a step with vaporization at Tvap. Is this enthalpy of vaporization larger or smaller in magnitude from the standard value?

For the reaction C(s,graphite) + 02(g) CO2(g) AG° = -394.3 kJ and AS° = 2.9 J/K at 290 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 290 K. The standard enthalpy change for the reaction of 2.38 moles of C(s,graphite) at this temperature would be kJ.