Why is the Fe-Fe3-C phase diagram a metastable phase diagram instead of a true equilibrium phase diagram? (b) Define the following phases that exist in the Fe-Fe3-C phase diagram: (i) austenite, (ii) ferrite, (iii) cementite, (iv) ferrite.
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Why is the Fe-Fe3-C phase diagram a metastable phase diagram instead of a true equilibrium phase diagram? (b) Define the following phases that exist in the Fe-Fe3-C phase diagram: (i) austenite, (ii) ferrite, (iii) cementite, (iv) ferrite.
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- Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).Part 1: (a) Calculate K at 298 K for the following reaction: 2C(graphite) + O₂(g) = 2CO(g) K = 1.25 C(graphite) CO(g) 0₂ (8) Part 2 out of 2 AGⓇ x 10 || 48 (b) Use the equilibrium constant to calculate AG (in kJ) at 298 K for the following reaction: 2HCl(g) + Br₂(1) ⇒ 2HBr(g) + Cl₂(g) K= 2.22 × 10-15 at 298 K AG KJ kJ C mol x 10 0 -137.2 0 kJ26. For each reaction, an equilibrium constant at 298 K is given. Calculate AG for each reaction. a) Br() + H,(g) 2 HBr(g) K,- 4.4 x 10" b) H,O(() = H,O(g) K,= 3.17 x 101 c) N;(g) + 3 Hạ(g) 2 NH,(g) K = 3,5 x 10
- Consider the equilibrium 21(8) — 12(8) Calculate the standard change in enthalpy, AH AH = 62.25 kJ/mol for l₂(g) and AH, 106.6 kJ/mol for I(g).Calculate K for the reaction Ag₂ Q(8) + CO(g) ⇒ 2 Ag(s) + CO₂ (g) given the following information: K = Ag₂ O(s) + H₂ (g) ≥ 2 Ag(s) + H₂O(g) K₁ = 1.01 X 10¹3 H₂ (g) + CO₂ (g) ⇒ H₂ O(g) + CO(g) K2 = 0.7714.
- Calculate AG° for H20(g) + 02(g) = H2O2 (g) at 595 K, using the following data: H2 (9) + O2 (9) = H2O2(g) K = 3.7 x 1037 at 595 K 2H2 (9) + O2 (9) = 2H2O(g) K = 2.6 × 10® at 595 K AG kJ/molThe phase diagram for SO2 is shown here. (a) What doesthis diagram tell you about the enthalpy change in thereaction SO2(l) ---->SO2(g)? (b) Calculate the equilibriumconstant for this reaction at 100 °C and at 0 °C.(c) Why is it not possible to calculate an equilibrium constantbetween the gas and liquid phases in the supercriticalregion? (d) At which of the three points marked in reddoes SO2(g) most closely approach ideal-gas behavior?(e) At which of the three red points does SO2(g) behaveleast ideally?2 Zn*2 (aq) +Fe*2 (ag) + 6 CN (ag) = Zn2Fe(CN)6 (5) [znz Fe (CN),] A certain equilibrium reaction has K = 2.8 x 10-13 at 25 °C. What is the change in Gibbs Free Energy (kJ/mole) under these conditions?
- Consider the following equilibrium: 4Ag(s) + O2(g) ---> 2Ag2O(s). At 298 K, the equilibrium constant for this reaction is K = 8.44 X 10^3. What is the change in Gf under standard state conditions for Ag2O(s)?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•KNaBr –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…18. From the following reacțions and thermal data at 25°C 3 2 Fe(s) +O:(g) 2 Fe,O:(s) ΔΗ - 196,500 cal %3D 1 2 FeO(s) + ; 0:(g) - Fe2O:(s) AH° = -67,900 cal Fe(s) + 2 H* (a = 1) = Fe++ (a = 1) + H2(g) AH° = -21,000 cal %3D %3D 1 H:(g) H* (a = 1) AHº = 0 cal %3D Hig) +0:(g) = H;0(1) AH° = -68,320 cal %3D calculate AH° of the reaction: FeO(s) + 2 H+ (a = 1) = H;0(1) + Fe++ (a = 1) %3D