Calculate the standard enthalpy change for the following neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given: - The standard enthalpy of formation (ΔHf°) of NaCl(aq) is -407.3 kJ/mol.- The standard enthalpy of formation (ΔHf°) of H2O(l) is -285.8 kJ/mol.- The specific heat capacity (c) of the solution is 4.18 J/g°C.- The temperature change (ΔT) during the reaction is 15°C. - The mass (m) of the solution is 50.0 g. What is the standard enthalpy change (ΔH°) for the neutralization reaction between HCl and NaOH?

Question

Calculate the standard enthalpy change for the following neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH):HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)Given: - The standard enthalpy of formation (ΔHf°) of NaCl(aq) is -407.3 kJ/mol.- The standard enthalpy of formation (ΔHf°) of H2O(l) is -285.8 kJ/mol.- The specific heat capacity (c) o

Answer 1

To calculate the standard enthalpy change  H  for the neutralization reaction between HCl and NaOH, we can use the following equation:H =  Hf products  -  Hf reactants First, let's identify the standard enthalpy of formation  Hf  for the reactants:- HCl aq  and NaOH aq  are strong acids and bases, so their Hf values are considered to be zero.Now, let's plug the values into the equation:H = [ -407.3 kJ/mol  +  -285.8 kJ/mol ] - [ 0 kJ/mol  +  0 kJ/mol ]H = -693.1 kJ/molThe standard enthalpy change  H  for the neutralization reaction between HCl and NaOH is -693.1 kJ/mol.