Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), given the appropriate thermochemical equations and the following information:- Heat evolved during the reaction, Q = -58.35 kJ/mol- Moles of HCl used, n(HCl) = 0.05 mol- Concentration of NaOH solution, c(NaOH) = 2 M- Volume of NaOH solution used, V(NaOH) = 25.0 mL (assuming density of 1.00 g/mL)Show your calculations and include units in your final answer.

Question

Calculate the standard enthalpy change for the neutralization reaction between hydrochloric acid (HCl) and sodium hydroxide (NaOH), given the appropriate thermochemical equations and the following information:- Heat evolved during the reaction, Q = -58.35 kJ/mol- Moles of HCl used, n(HCl) = 0.05 mol- Concentration of NaOH solution, c(NaOH) = 2 M- Volume of NaOH solution used, V(Na

Answer 1

First, we need to find the moles of NaOH used in the reaction. We can do this using the concentration and volume of the NaOH solution:moles of NaOH = c NaOH   V NaOH Since the volume is given in mL, we need to convert it to L:V NaOH  = 25.0 mL   1 L / 1000 mL  = 0.025 LNow we can find the moles of NaOH:moles of NaOH = 2 M  0.025 L = 0.05 molSince the reaction between HCl and NaOH is a 1:1 reaction, the moles of HCl and NaOH are equal. Therefore, the moles of HCl used in the reaction are also 0.05 mol.Now we can calculate the standard enthalpy change for the neutralization reaction  H  using the heat evolved  Q  and the moles of HCl  n HCl  :H = Q / n HCl H = -58.35 kJ/mol / 0.05 molH = -1167 kJ/molThe standard enthalpy change for the neutralization reaction between hydrochloric acid and sodium hydroxide is -1167 kJ/mol.