Consider the dissolution of NaBr and NaI. The values provided here will be helpful for answering the following questions.             ΔH∘solnΔHsoln∘ (kJ/mol)              ΔS∘solnΔSsoln∘ Jmol•K​NaBr            –0.860                                  57.0NaI              –7.50                                     74.0   Write a balanced equilibrium equation for the dissolution of NaI in water. Include phases.       Which of the following explains why the entropy change is greater for the dissolution of NaI compared to the dissolution of NaBr?Choose one: A.  The interactions between bromide ions with other bromide ions is stronger than the interactions between iodide ions with other iodide ions. B.  The cation forms stronger ion-dipole networks with water in NaBr than NaI because of the weaker bond to Br.C.  The more negative change in enthalpy observed with NaI implies greater dissociation and hence greater entropy.D.  Iodide has weaker ion-dipole interactions with water than bromide. E.  The bromide ion has a more negative charge than the iodide ion. Therefore, because of the greater charge, it forms a stronger ion-dipole network with water.    Calculate the change in free energy if 1.02 moles of NaI is dissolved in water at 25.0°C.     kJ      What is the dissolution of 1.00 mol of NaBr at 298.15 K?

A student determines the value of the equilibrium constant to be 2.69×10-10 for the following reaction.Fe2O3(s) + 3H2(g)2Fe(s) + 3H2O(g)Based on this value of Keq:G° for this reaction is expected to be (greater, less) fill in the blank than zero.Calculate the free energy change for the reaction of 2.05 moles of Fe2O3(s) at standard conditions at 298K.     G°rxn =  ____kJ

A student determines the value of the equilibrium constant to be 4.09×10-18 for the following reaction. Fe;O4(s) + 4H2(g)→3FE(s) + 4H,O(g) Based on this value of Kegi AG° for this reaction is expected to be (greater, less) greater than zero. Calculate the free energy change for the reaction of 1.76 moles of Fe;0,(s) at standard conditions at 298K. AG°. rxn = 190.5 kJ

Use the following information to calculate the standard free energy change (AG°) and equilibrium constant (K) for the following system: CaCO3(s) = CaO(s) + CO2(g) T= 900.0 °C AH;° values (kJ/mol): CaO(s) = -634.9, CO2(g) = -393.51, CaCO3(s) =–1220.0 %3D S° values (J/mol K): CaO(s) = 110.0, CO2(g) = 213.8, CACO3(s) = 38.1 %3D %3D %3D

Calcium ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) Ca (s) 0 0 41.4 Ca (g) 178.2 144.3 158.9 Ca2+ (g) 1925.9     CaC2 (s) -59.8 -64.9 70.0 CaCO3 (s, calcite) -1206.9 -1128.8 92.9 CaCl2 (s) -795.8 -748.1 104.6 CaF2 (s) -1219.6 -1167.3 68.9 CaH2 (s) -186.2 -147.2 42.0 CaO (s) -635.1 -604.0 39.8 CaS (s) -482.4 -477.4 56.5 Ca(OH)2 (s) -986.1 -898.5 83.4 Ca(OH)2 (aq) -1002.8 -868.1 -74.5 Ca3(PO4)2 (s) -4126.0 -3890.0 241.0 CaSO4 (s) -1434.1 -1321.8 106.7 CaSiO3 (s) -1630.0 -1550.0 84.0       Carbon ΔHof (kJ/mol) ΔGof (kJ/mol) So (J/mol K) C (s, graphite) 0 0 5.7 C (s, diamond) 1.9 2.9 2.4 C (g) 716.7 671.3 158.1 CCl4 (l) -135.4 -65.2 216.4 CCl4 (g) -102.9 -60.6 309.9 CHCl3 (l) -134.5 -73.7 201.7 CHCl3 (g) -103.1 -70.3 295.7 CH4 (g) -74.8 -50.7 186.3 CH3OH (g)…

Consider the oxidation of a generic metal M into a metal oxide as shown in the reaction. 3 2 M(s) + 0,(g) → M,O,(s) For this reaction, AHan -4.811 kJ/mol and ASirn = 12.06 J/(mol·K) at 298 K. What is the standard change in Gibbs free energy for the reaction in the forward direction? AGixn kJ/mol What is the equilibrium constant of this reaction at 298 K? Kp What is the equilibrium pressure of O,(g) over M(s) at 298 K? Ро, atm

[Review Topics] [References] Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH° = -111 kJ/mol, and K. = 0.159 , at 723 K. N2 (g) + 3 H, (g)=2 NH3 (g) When 0.21 moles of NH, (g) are added to the equilibrium system at constant temperature: The value of K The value of Qe K. The reaction must O run in the forward direction to restablish cquilibrium. O run in the reverse direction to restablish coquilibrium. O remain the same. It is already at cquilibrium. The concentration of H, will Submit Answer Retry Entire Group 1 more group attempt remaining

11. What is the ΔHº-value for the dissolution of NaOH (s) to yield 1.00... m NaOH (aq), given:      Na (s) + 1/2 O2 (g) + 1/2 H2 (g) → NaOH (s)              ΔHfº = -425.9 kJ/mol   Na (s) + 1/2 O2 (g) + 1/2 H2 (g) → NaOH (aq, 1 m)    ΔHfº = -469.2 kJ/mol     Multiple Choice   -895.1 kJ/mol   -43.3 kJ/mol   +1.102 kJ/mol   +43.3 kJ/mol   +895.1 kJ/mol