Question

Calculate the standard enthalpy change for the reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) given the following enthalpy changes: ∆Hf° for NaCl(aq) = -407.3 kJ/mol ∆Hf° for H2O(l) = -285.8 kJ/mol ∆Hf° for NaOH(aq) = -469.11 kJ/mol ∆Hf° for HCl(aq) = -167.16 kJ/mol

Answer 1

To calculate the standard enthalpy change for the 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 the individual steps. In this case, we can use the enthalpy changes of formation  Hf  for the reactants and products.The equation for calculating the standard enthalpy change for the reaction is:H =  [Hf products ] -  [Hf reactants ]For this reaction, the products are NaCl aq  and H2O l , and the reactants are NaOH aq  and HCl aq . Plugging in the given values, we get:H = [ -407.3 kJ/mol  +  -285.8 kJ/mol ] - [ -469.11 kJ/mol  +  -167.16 kJ/mol ]H =  -407.3 - 285.8 + 469.11 + 167.16  kJ/molH =  -693.1 + 636.27  kJ/molH = -56.83 kJ/molThe standard enthalpy change for the reaction is -56.83 kJ/mol.