Practice ProblemATTEMPT
Will the following reaction occur spontaneously at
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- Consider the following reaction at 298K.Cu2+ (aq) + Cr2+ (aq) Cu+ (aq) + Cr3+ (aq)Which of the following statements are correct?Choose all that apply. Eocell > 0 The reaction is reactant-favored. n = 4 mol electrons delta Go > 0 K > 1 Submit Answerarrow_forwardWhat is ΔH, ΔS and ΔG° at 1000 °C for the following reaction? CaCO3 (s) = CaO (s) + CO2 (g) ΔHf° (KJ) S° (J/K CaCO3 -1206.9 92.9 CaO -635.1 38.2 CO2 -393.5 213.7 ΔHAnswerKJ ΔSAnswerJ/K ΔGoAnswerKJ (b) Is the reaction spontaneous at 1000 °C and 1 atm? Answer (c) What is the value of Kp at 1000 °C for this reaction given that (R = 8.314 J/K mol) KpAnswer (d) What is the partial pressure of CO2 (g)?Answeratmarrow_forwardploA plot of ln K (equilibrium constant) for a reaction vs 1/T gave a line with slope - 2040 kelvin and a y-intercept of 2.41. What is the standard entropy change for the reaction?arrow_forward
- Predict the sign of ΔS°, if possible, for the following reaction. A. N2(g) + O2(g) ----> 2NO(g) B. C2H2(g) + 2H2(g) ----> C2H6(g) Question 3 options: positive negative not predictablearrow_forwardConsider the reactionNH4Cl(aq)------->NH3(g) + HCl(aq)Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.ANSWER:arrow_forwardIn the gas-phase reaction A + B = C + 2 D, it was found that, when 2.00 mol A, 1.00 mol B and 3.00 mol D were mixed and allowed to come to equilibrium at 25oC, the resulting mixture contained 0.79 mol C at a total pressure of 1.00 bar. Calculate Kp. Answer: [.960]arrow_forward
- What is K if ΔG° = -20.0 kJ for a reaction at 25°? (SHOW CALCULATION)arrow_forwardConsider the reactionCa(OH)2(aq) + 2 HCl(aq) ----> CaCl2(s) + 2 H2O(l)Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K.ANSWER: ____________arrow_forwardThe concentration equilibrium constant Kc for the reaction 2 SO3 (g) = 2 SO2 (g) + O2 (g) is 0.0271 mol dm-3 at 1100 K. Calculate Kp at that temperature. Answer: [2.48 bar]arrow_forward
- The standard free-energy change for the Haber process at 25 °C was obtained in Sample Exercise 19.9 for the Haber reaction:N21g2 + 3 H21g2 ∆ 2 NH31g2 ∆G° = -33.3 kJ>mol = -33,300 J>mol Use this value of ∆G° to calculate the equilibrium constant for the process at 25 °C.arrow_forwardExplain what happens as a reaction starts with ΔG < 0 (negative) and reaches the point where ΔG = 0.arrow_forwardIn the soilution below why is the value of K 6.7478 and not 6.7478 * 1015 Question Calculate ΔG° and K (at 298K) for this reaction: 2H2S(g) + SO2(g) ↔ 3S(s) + 2H2O(g) Expert Answer Step 1 The standard Gibbs free energy can be calculated as follows: ∆G∘=nΣ∆G∘(Products)−mΣ∆G∘(Reactants) where ΔGo: Standard Gibbs free energy n and m are stoichiometric coefficients and Equilibrium Constant (K) can be calculated as follows: ∆G∘=−RTlnK where R: Gas Constant (8.314 JK-1mol-1) ΔGo: Standard Gibbs free energy K: Equilibrium Constant T: Temperature (298 K) Step 2 The standard Gibbs free energy of formation for the following substances are: ΔGof (H2S) = -33.4 kJ/mol ΔGof (SO2) = -300.1 kJ/mol ΔGof (S) = 0 kJ/mol ΔGof (H2O) = -228.6 kJ/mol Substituting all these values in this equation- ∆G∘======[3∆Gf∘(S)+2∆Gf∘(H2O)]−[2∆Gf∘(H2S)+1∆Gf∘(SO2)][3(0 kJ/mol)+2(−228.6 kJ/mol)]−[2(−33.4 kJ/mol)+1(−300.1 kJ/mol)][0 −457.2] kJ/mol−[−66.8−300.1] kJ/mol−457.2−[−366.9]…arrow_forward
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