a.) Calculate the equilibrium constant for the following reaction at 298.15 K from cell potential data. Express the answer as lnK.
Sn4+ + 2Fe2+ ----> Sn2+ + 2Fe3+
b.) Calculate the standard Gibbs free energy change in kJ/mol at 298.15 K for the following reaction from cell potential data:
3Sn4+ + 2Cr ----> 3Sn2+ + 2Cr3+
a.) Calculate the equilibrium constant for the following reaction at 298.15 K from cell potential data....
A.) Calculate ∆G0 for a reaction with K = 5,219.2 at 298.15 K. Report your answer in kJ/mol. B.) Calculate ∆G0 for a reaction with K = 0.05964 at 298.15 K. Report your answer in kJ/mol. C.) Calculate the equilibrium constant at 298.15 K for a reaction with ∆G0 = -12.26 kJ/mol. D.) Calculate the equilibrium constant at 298.15 K for a reaction that has a ∆G0 = 14.18 kJ/mol.
Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 641 K for the following reaction. NH4Cl(s) NH3(g) + HCl(g) ΔΗ /kJ mol-1 -314.4 -45.9 -92.3 Smº /JK-mol-1 94.6 192.8 186.9 Cp.m /JK-mol-1 84.1 35.1 29.1 Do not use the Van't Hoff equation, In(K/K) = -(AHR/R) (T2-1-T1-1) The value of the thermodynamic equilibrium constant is Kp = Number
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From the information in the Data section of the textbook, calculate the equilibrium constant at 338 K for the reaction: assuming that the reaction enthalpy is independent of temperature. Answer: 0.0835 The degree of dissociation, a, for the following reaction is 0.655 at 298 K and 1.00 bar total pressure. Find K Answer: 300 Given K 9.18E-8 for the reaction: laqi and the...
Use the data below, for 298.15 K, to calculate the thermodynamic equilibrium constant, kp, at 839 K for the following reaction. NH4Cl(s) NH3(g) + HCl(g) ΔΗ 7 kJ mol-1 -314.4 -45.9 -92.3 Smº JK-1 mol-1 94.6 192.8 186.9 Cp,m 84.1 35.1 29.1 /JK-1 mol-1 Do not use the Van't Hoff equation, In(K /K1) = -(AHR/R) (T2-1 - 7,-1) The value of the thermodynamic equilibrium constant is Kp= 7.6e14
Calculate the equilibrium constant at 298.15 K for each of the following reactions from the value of ΔG° given. Please keep 2 significant figures. (a) H2(g) + 2 Na(s) → 2 NaH(s) ΔG° = −66.9 kJ/mol K= (b) 4 NO2(g) + O2(g) → 2 N2O5(g) ΔG° = 29.2 kJ/mol K= (c) SiO2(s) + 2 Cl2(g) → SiCl4(l) + O2(g) ΔG° = 236.5 kJ/mol K= †
What is the equilibrium constant of the formation reaction of ammonia gas at 298.15 K under the conditions that the partial pressures of N2, H2 and NH3 are 2.00 bar, 3.00 bar, and 2.00 bar, respectively? (Hint: Be careful! This is a trick question! The equilibrium constant is what it is. It is not dependent on what the partial pressures happen to be at any particular time. K is purely dependent on the standard Gibbs free energy of the reaction..)
Using standard thermodynamic data (linked), calculate the equilibrium constant at 298.15 K for the following reaction. HCl(g) + NH3(g)--->NH4Cl(s) K =
Calculate the standard Gibbs energy of the following reaction at T = 298.15 K using the data given below. 2 CH3CHO(g) + O2(g) ® 2 CH3COOH(l) Som(CH3COOH(l)) = 159.8 J K–1 mol–1 DfHo(CH3COOH(l)) = –484.5 kJ mol–1 Som(CH3CHO(g)) = 250.3 J K–1 mol–1 DfHo(CH3CHO(g)) = –166.19 kJ mol–1 Som(O2(g)) = 205.14 J K–1 mol–1
The equilibrium constant, K, for a redox reaction is related to the standard potential, E∘, by the equation lnK=nFE∘RT where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e−) , R (the gas constant) is equal to 8.314 J/(mol⋅K) , and T is the Kelvin temperature. Calculate the standard cell potential (E∘) for the reaction X(s)+Y+(aq)→X+(aq)+Y(s) if K = 5.51×10−3.
Use tabulated half-cell potentials (found in appendix IV-D) to calculate Eºcell and then calculate AGºrxn for the following reaction at 25 °C. 2Fe3+ (aq) + 3Sn(s) → 2Fe(s) + 3Sn2+ (aq) Hint: the standard potential of Fe3+ + 3e + Fe(s) is -0.036 and the standard potential of Sn2+ + 2e → Sn(s) is -0.140. AG° = -5 kJ/mol AG° = -10 kJ/mol AG° = -20 kJ/mol AG° = -30 kJ/mol