0 votes
30 views
ago in Chemical thermodynamics by (450 points)
Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), given that the reaction produces water and sodium chloride as products. The molar enthalpy of formation for water is -285.8 kJ/mol and for NaCl is -411.2 kJ/mol. The balanced chemical equation for the reaction is: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

1 Answer

0 votes
ago by (490 points)
To calculate the standard enthalpy change for the neutralization reaction, we can use Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step in the reaction. In this case, we can consider the formation of the products from their constituent elements, and then subtract the enthalpy of formation for the reactants.The balanced chemical equation for the reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l The enthalpy change for the reaction  H  can be calculated using the following equation:H = H products  - H reactants For the products, we have:1 mol of NaCl aq  with an enthalpy of formation of -411.2 kJ/mol1 mol of H2O l  with an enthalpy of formation of -285.8 kJ/molFor the reactants, we have:1 mol of HCl aq  with an enthalpy of formation of -92.3 kJ/mol  from reference data 1 mol of NaOH aq  with an enthalpy of formation of -469.6 kJ/mol  from reference data Now, we can plug these values into the equation:H = [ -411.2 kJ/mol  +  -285.8 kJ/mol ] - [ -92.3 kJ/mol  +  -469.6 kJ/mol ]H =  -696.0 kJ/mol  -  -561.9 kJ/mol H = -134.1 kJ/molTherefore, the standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -134.1 kJ/mol.

Related questions

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