Question

Calculate the standard enthalpy change for the reaction between 25 mL of 0.1 M HCl and 25 mL of 0.1 M NaOH, assuming the reaction takes place in aqueous solution and is complete. The temperature of the solutions should be maintained at a constant 25°C during the reaction. (Assume the density of the solutions is 1 g/mL and the specific heat capacity of the solutions is 4.18 J/g°C).

Answer 1

To calculate the standard enthalpy change for the reaction between 25 mL of 0.1 M HCl and 25 mL of 0.1 M NaOH, we first need to write the balanced chemical equation for the reaction:HCl aq  + NaOH aq   NaCl aq  + H2O l Next, we need to determine the moles of reactants involved in the reaction. Since both solutions have the same concentration and volume, they will react in a 1:1 ratio, and the limiting reactant will be completely consumed.moles of HCl = moles of NaOH = 0.1 M  0.025 L = 0.0025 molesNow, we need to find the standard enthalpy change  H  for the reaction. The standard enthalpy change for the reaction between HCl and NaOH is -55.9 kJ/mol. This value is negative because the reaction is exothermic, meaning it releases heat.To find the enthalpy change for the specific amount of reactants we have, we can use the following equation:H = moles  standard enthalpy change per moleH = 0.0025 moles   -55.9 kJ/mol  = -0.13975 kJSo, the standard enthalpy change for the reaction between 25 mL of 0.1 M HCl and 25 mL of 0.1 M NaOH is -0.13975 kJ.