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- What 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_forwardFor which of the following is the change in entropy positive? Check all that apply. Explain. A. C 6 H 12 (l) C 6 H 12 (g)B. Dilution of 1.0 mL of 1.0 M NaOH with 90 mL of waterC. . 2 O 2 (g) + 2 SO (g) 2 SO 3 (g)D. grinding a large crystal of NaCl to powderE. Cooling water from room temperature to 15°C.arrow_forwardCalculate K at 298 K for the following reaction: NO(g) + 1/2 O2 (g) <-> NO2 (g) Delta G (kJ/mol) for the following: NO(g) 86.60, NO2(g) 51, O2(g) 0 (answer in scientific notation)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_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picturearrow_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picturePLEASE EXPLAIN ALSO PLSarrow_forward
- Quick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picture/ i think it is to be answered in paragraph form .arrow_forwardQuick overview of our lesson: Our topic is all about Second Law of Thermodynamics. Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left. please do help me with the questions on the picture/ PLEASE ANSWER DIRECTLY WITH THE EXPLANATION.arrow_forwardQuick overview of our lesson:Our topic is all about Second Law of Thermodynamics.Gibbs’ free energy, G is defined by G = H – TSwhere H is the enthalpy, T is the temperature (in Kelvins), and S is the entropy. In a chemical reaction,R ↔P (R are reactants and P are products) at aconstant temperature we have ∆G = ∆H – T∆S.If ∆G < 0 the reaction may proceed spontaneously tothe right.If ∆G = 0 the reaction is in equilibrium.If ∆G > 0 the reaction may proceed spontaneously to the left.please do help me with the questions on the picturearrow_forward
- Consider the reaction Fe(s) + 2 HCl(aq) → FeCl2(s) + H2(g) Using the standard thermodynamic data in the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. ANSWER: _______arrow_forwardConsider the reactionCH4(g) + H2O(g)->3H2(g) + CO(g)Using the standard thermodynamic data in the tables linked above, calculate deltaG for this reaction at 298.15K if the pressure of each gas is 21.37 mm Hg.ANSWER: kJ/molarrow_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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