0 votes
25 views
ago in Chemical thermodynamics by (48.8k points)
What is the standard enthalpy change for the neutralization reaction of hydrochloric acid (HCl) and sodium hydroxide (NaOH) at 25°C and 1 atm pressure? The balanced equation for the reaction is: HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

1 Answer

0 votes
ago by (48.8k points)
The standard enthalpy change for the neutralization reaction of hydrochloric acid  HCl  and sodium hydroxide  NaOH  at 25C and 1 atm pressure can be determined using the standard enthalpy of formation values for the reactants and products.The balanced equation for the reaction is: HCl  aq  + NaOH  aq   NaCl  aq  + H2O  l Using the standard enthalpy of formation values  Hf  at 25C:Hf  HCl, aq  = -167.2 kJ/molHf  NaOH, aq  = -469.15 kJ/molHf  NaCl, aq  = -407.27 kJ/molHf  H2O, l  = -285.83 kJ/molThe standard enthalpy change for the reaction  H  can be calculated using Hess's law:H =  [Hf  products ] -  [Hf  reactants ]H = [ -407.27 kJ/mol  +  -285.83 kJ/mol ] - [ -167.2 kJ/mol  +  -469.15 kJ/mol ]H =  -693.1 kJ/mol  -  -636.35 kJ/mol H = -56.75 kJ/molTherefore, the standard enthalpy change for the neutralization reaction of HCl and NaOH at 25C and 1 atm pressure is -56.75 kJ/mol.

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...