Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), given the balanced chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l). The enthalpies of formation for NaCl(aq) and H2O(l) are -407.3 kJ/mol and -285.8 kJ/mol, respectively. The specific heat capacity of the solution is 4.18 J/(g*K), and the mass of the resulting solution is 100 g.

Question

Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), given the balanced chemical equation: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l). The enthalpies of formation for NaCl(aq) and H2O(l) are -407.3 kJ/mol and -285.8 kJ/mol, respectively. The specific heat capacity of the solution is 4.18 J/(g*K), and the mass of th

Answer 1

To calculate the standard enthalpy change for the neutralization reaction, we can use the following equation:H_reaction =  H_f products  -  H_f reactants The enthalpies of formation for the products are given as -407.3 kJ/mol for NaCl aq  and -285.8 kJ/mol for H2O l . Since HCl aq  and NaOH aq  are strong acids and bases, their enthalpies of formation can be assumed to be zero.H_reaction = [ -407.3 kJ/mol  +  -285.8 kJ/mol ] - [0 + 0]H_reaction = -693.1 kJ/molThe standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -693.1 kJ/mol.