Question

Calculate the standard enthalpy change (∆H°) for the reaction between 25.0 mL of 0.100 M HCl and 25.0 mL of 0.100 M NaOH at 25°C. The density of each solution is 1.00 g/mL and the heat capacity of the solution is 4.18 J/g·°C. The reaction takes place in a calorimeter, and the temperature increase is measured to be 8.47°C.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we first need to determine the heat  q  released or absorbed during the reaction using the formula:q = mcTwhere m is the mass of the solution, c is the heat capacity, and T is the change in temperature.Since the density of each solution is 1.00 g/mL, the mass of the solution can be calculated as follows:mass = volume x densitymass =  25.0 mL HCl + 25.0 mL NaOH  x 1.00 g/mLmass = 50.0 gNow we can calculate the heat  q  released or absorbed during the reaction:q =  50.0 g  x  4.18 J/gC  x  8.47C q = 1774.53 JNext, we need to determine the moles of the limiting reactant. Since both HCl and NaOH have the same concentration and volume, they will react in a 1:1 ratio, and either one can be considered the limiting reactant. We will use HCl for our calculation:moles HCl =  0.100 mol/L  x  0.025 L moles HCl = 0.0025 molNow we can calculate the standard enthalpy change  H  per mole of the reaction:H = q / moles HClH = 1774.53 J / 0.0025 molH = 70981.2 J/molSince the standard enthalpy change is usually expressed in kJ/mol, we can convert the value:H = 70.98 kJ/molTherefore, the standard enthalpy change for the reaction between 25.0 mL of 0.100 M HCl and 25.0 mL of 0.100 M NaOH at 25C is -70.98 kJ/mol  the negative sign indicates that the reaction is exothermic .