Please ignore my writing, but could you explain how to find the answer? Will leave good rating! Thank you! R²1 = nonspontaneous(+)75(-) (-)-RT In(K)Practice ProblemAG°creating charged particles.Consider the base-ionization reaction for ammonia, NH3.Calculate the AG for this reaction using the fact that the Kc for+NH₂* reaction is 5.6 × 10-10. Assume at standard temp26.57 -237.1-79.37157.3NH3(aq) + H₂O(l) = NH4+ (aq) + OH(aq) we dont krbase be accepts protonsentropy.AG, (kJ/mol)at 25 °CAG = 27.1 kJ/molSubstanceNH3(aq)-26.57NHalog)-79.37Using the AG, from the table:H₂O(l)-237.1AG = 27.0 kJ/molOH (aq)-157.3Need enthalpy to determine endo exo-

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75 (+) (-) (-) -RT In(K) Practice Problem AG° creating charged particles. Consider the base-ionization reaction for ammonia, NH3. Calculate the AG for this reaction using the fact that the Kc for NH₁* reaction is 5.6 × 10-10. Assume at standard temp 26.57 -237.1 -79.37 NH₂(aq) + H₂O(l) = base be accepts protons -157.3 NH4+ (aq) + OH(aq) we don't entropy. AG, (kJ/mol) AG° = 27.1 kJ/mol Substance at 25 °C NH3(aq) -26.57 Nha (09) -79.37 H₂O(l) -237.1 Using the AG, from the table: AG = 27.0 kJ/mol OH (aq) -157.3 1

75 (+) (-) (-) -RT In(K) Practice Problem AG° creating charged particles. Consider the base-ionization reaction for ammonia, NH3. Calculate the AG for this reaction using the fact that the Kc for NH₁* reaction is 5.6 × 10-10. Assume at standard temp 26.57 -237.1 -79.37 NH₂(aq) + H₂O(l) = base be accepts protons -157.3 NH4+ (aq) + OH(aq) we don't entropy. AG, (kJ/mol) AG° = 27.1 kJ/mol Substance at 25 °C NH3(aq) -26.57 Nha (09) -79.37 H₂O(l) -237.1 Using the AG, from the table: AG = 27.0 kJ/mol OH (aq) -157.3 1

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

Chapter18: Electrochemistry

Section: Chapter Questions

Problem 9RQ: What characterizes an electrolytic cell? What is an ampere? When the current applied to an...

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