Solution Found!
At 25°C, the standard enthalpy of formation of HF(aq) is
Chapter 6, Problem 118P(choose chapter or problem)
At \(25^{\circ} \mathrm{C}\), the standard enthalpy of formation of HF(aq) is given by - 320.1 kJ/mol; of \(O H^{-}(a q)\), it is -229.6 kJ/mol; of \(F^{-}(a q)\), it is - 329.1 kJ/mol; and of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), it is -285.8 kJ/mol.
(a) Calculate the standard enthalpy of neutralization of HF(aq):
\(H F(a q)+O H^{-}(a q) \rightarrow H_{2} O(l)\)
(b) Using the value of -56.2 kJ as the standard enthalpy change for the reaction
\(H^{+}(a q)+O H^{-}(a q) \rightarrow H_{2} O(l)\)
calculate the standard enthalpy change for the reaction
\(H F(a q) \rightarrow H^{+}(a q)+F^{-}(a q)\)
Questions & Answers
QUESTION:
At \(25^{\circ} \mathrm{C}\), the standard enthalpy of formation of HF(aq) is given by - 320.1 kJ/mol; of \(O H^{-}(a q)\), it is -229.6 kJ/mol; of \(F^{-}(a q)\), it is - 329.1 kJ/mol; and of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), it is -285.8 kJ/mol.
(a) Calculate the standard enthalpy of neutralization of HF(aq):
\(H F(a q)+O H^{-}(a q) \rightarrow H_{2} O(l)\)
(b) Using the value of -56.2 kJ as the standard enthalpy change for the reaction
\(H^{+}(a q)+O H^{-}(a q) \rightarrow H_{2} O(l)\)
calculate the standard enthalpy change for the reaction
\(H F(a q) \rightarrow H^{+}(a q)+F^{-}(a q)\)
ANSWER:Step 1 of 3
(a) Here we have to calculate the standard enthalpy of neutralization of HF(aq) ().
Given:
The given chemical reaction is ,
Temperature = 25 °C
Enthalpy of formation of HF = -320.1 KJ/mol
Enthalpy of formation of 
Enthalpy of formation of 
Enthalpy of formation of water = -285.8 KJ/mol