**Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K.**Reaction: \[ \text{1A + 3B} \rightarrow \text{5C + 4D + 3E} \]| Substance | \(\Delta H_f^0\) (kJ/mol) | \(S^0\) (J/mol/K) ||-----------|-----------------|----------------|| A | -19.68 | 13.41 || B | -9.29 | 10.92 || C | 0.83 | 19.81 || D | 15.35 | 9.48 || E | 3.32 | 0.15 |**Remember,** \[ \text{pK} = -\log(K) \]**Report your answer to two decimal points.**---**Explanation:**1. **\(\Delta H_f^0\) column**: Lists the standard enthalpy of formation for each substance in kJ/mol.2. **\(S^0\) column**: Lists the standard entropy for each substance in J/mol/K.To calculate the equilibrium constant (K), use the Gibbs free energy change (\(\Delta G^0\)) for the reaction:\[ \Delta G^0 = \Delta H^0 - T\Delta S^0 \]Where:- \(\Delta H^0\) and \(\Delta S^0\) are determined from the reaction data.- \(T\) is the temperature in Kelvin (298 K for 25°C).Calculate \(\Delta G^0\) and then use the relation:\[ \Delta G^0 = -RT \ln(K) \]to solve for \(K\) and subsequently, \(\text{pK} = -\log(K)\).

The difference between the initial-state energy and the equilibrium-state energy is known as free energy of a system.This is the amount of energy available for doing external work. At constant temperature and pressure relationship between enthalpy of the reaction, entropy of the reaction and free energy of reaction is given by: ∆G =∆H- T∆S Given reaction: 1A + 3B →5C +4D+ 3E Given information: Temperature of the reaction = 250C = 25 +273K = 298 K Enthalpy of formation ∆Hf0 of A = - 19.68 kJ/mol Enthalpy of formation ∆Hf0 of B = -9.29 kJ/mol Enthalpy of formation ∆Hf0 of C = 10.83 kJ/mol Enthalpy of formation ∆Hf0 of D = 15.35 kJ/mol Enthalpy of formation ∆Hf0 of E = 3.32 kJ/mol Entropy of formation ∆Sf0of A = 13.41 J/mol K Entropy of formation ∆Sf0 of B = 10.92 J/mol K Entropy of formation ∆Sf0 of C = 19.81 J/mol K Entropy of formation ∆Sf0 of D = 9.48 J/mol K Entropy of formation ∆Sf0of E = 0.15 J/mol KEnthalpy of the reaction is given by: Enthalpy of the reaction= Enthalpy of formation of product -Enthalpy of formation of reactant ∆Hr =5×∆Hf+4×∆Hf+3×∆Hf -3×∆Hf-1×∆Hf ∆Hr =5×0.83+4×15.35+3×3.32 -3×-9.29-1×-19.68 =123.06 kJ/mol 1kJ =1000J123.06 kJ/mol =123.06 ×1000 J/mol = 123060 J/mol Therefore, the enthalpy of the reaction is 123060 kJ/mol. Entropy of the reaction is given by: Entropy of the reaction= Entropy of formation of product -Entropy of formation of reactant ∆Sr =5×∆Sf+4×∆Sf+3×∆Sf -3×∆Sf-1×∆Sf ∆Sr =5×19.81+4×9.48+3×0.15 -3×10.92-1×13.41 =91.25J/K mol Therefore, entropy of the reaction is 91.25 J/K molFrom thermodynamic equation: ∆G =∆H- T∆S Substitute, 123060 J/mol enthalpy and 91.25 J/K mol entropy in the expression: ∆G =123060J/mol - 298×91.25 J/mol =95,867.5 J/mol From thermodynamic equation: ∆G =2.303 ×RT ×pK pK = ∆G2.303×R×T =95,867.52.303×8.314×298 ≈16.80 Therefore, pk of the reaction is 16.80.

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Chemistry

Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K. %3D 1A +3B →5C + 4D +3E AHF°(kJ/mol) A -19.68 s°(J/mol/K) 13.41 B -9.29 10.92 C 0.83 19.81 D 15.35 9.48 E 3.32 0.15 Remember, pK = -log(K) Report your answer to two decimal points.

Calculate the pK at 25 degrees C of the following reaction using the data in the table. Use R = 8.314 J/mol/K. %3D 1A +3B →5C + 4D +3E AHF°(kJ/mol) A -19.68 s°(J/mol/K) 13.41 B -9.29 10.92 C 0.83 19.81 D 15.35 9.48 E 3.32 0.15 Remember, pK = -log(K) Report your answer to two decimal points.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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