At 25°C, the standard enthalpy of formation of HF(aq) is -319.9 kJ/mol. Of OH (aq), it is -229.5 kJ/mol. Of F (aq), it is -330.3 kJ/mol, and of H20(1), it is -285.2 kJ/mol. (a) Calculate the standard enthalpy of neutralization of HF(aq): HF(aq) + OH (aq) F (aq) + H,0(1) kJ/mol (b) Using the value of-56.2 kJ as the standard enthalpy change for the reaction H*(aq) + OH (aq → H20(1) calculate the standard enthalpy change for the reaction HF(aq) H*(aq) +F (aq) kJ/mol
At 25°C, the standard enthalpy of formation of HF(aq) is -319.9 kJ/mol. Of OH (aq), it is -229.5 kJ/mol. Of F (aq), it is -330.3 kJ/mol, and of H20(1), it is -285.2 kJ/mol. (a) Calculate the standard enthalpy of neutralization of HF(aq): HF(aq) + OH (aq) F (aq) + H,0(1) kJ/mol (b) Using the value of-56.2 kJ as the standard enthalpy change for the reaction H*(aq) + OH (aq → H20(1) calculate the standard enthalpy change for the reaction HF(aq) H*(aq) +F (aq) kJ/mol
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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![At 25°C, the standard enthalpy of formation of HF(aq) is -319.9 kJ/mol. Of OH (aq), it is
-229.5 kJ/mol. Of F (aq), it is -330.3 kJ/mol, and of H20(1), it is -285.2 kJ/mol.
(a) Calculate the standard enthalpy of neutralization of HF(aq):
HF(aq) + OH¯(aq)F (aq) + H2O(1)
kJ/mol
(b) Using the value of-56.2 kJ as the standard enthalpy change for the reaction
H*(aq) + OH (aq → H2O(1)
calculate the standard enthalpy change for the reaction
HF(aq) → H*(aq) + F (aq)
kJ/mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff99cf1dd-3a0d-4998-93a2-f72f60d25195%2Fe7731dbb-80b1-437a-ae5f-b83aa5381cfe%2Fyijd8y_processed.jpeg&w=3840&q=75)
Transcribed Image Text:At 25°C, the standard enthalpy of formation of HF(aq) is -319.9 kJ/mol. Of OH (aq), it is
-229.5 kJ/mol. Of F (aq), it is -330.3 kJ/mol, and of H20(1), it is -285.2 kJ/mol.
(a) Calculate the standard enthalpy of neutralization of HF(aq):
HF(aq) + OH¯(aq)F (aq) + H2O(1)
kJ/mol
(b) Using the value of-56.2 kJ as the standard enthalpy change for the reaction
H*(aq) + OH (aq → H2O(1)
calculate the standard enthalpy change for the reaction
HF(aq) → H*(aq) + F (aq)
kJ/mol
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