Question

A chemistry student wants to determine the enthalpy change of a chemical reaction. They have 25 g of sodium hydroxide and want to mix it with 50 mL of 1.0 M hydrochloric acid. What is the maximum temperature change they can expect to see, assuming constant pressure and perfect heat transfer, and what is the enthalpy change of the reaction?

Answer 1

To determine the enthalpy change of the reaction, we first need to write the balanced chemical equation for the reaction between sodium hydroxide  NaOH  and hydrochloric acid  HCl :NaOH  aq  + HCl  aq   NaCl  aq  + H2O  l Next, we need to determine the limiting reactant. We have 25 g of NaOH and 50 mL of 1.0 M HCl. Let's convert these amounts to moles:Moles of NaOH = mass / molar mass = 25 g /  22.99 g/mol + 15.999 g/mol + 1.00784 g/mol  = 25 g / 39.997 g/mol  0.625 molMoles of HCl = volume  concentration = 0.050 L  1.0 mol/L = 0.050 molSince there are fewer moles of HCl, it is the limiting reactant.The enthalpy change for the reaction can be found using the standard enthalpy of formation values for the reactants and products. The standard enthalpy of formation for NaOH  aq , HCl  aq , NaCl  aq , and H2O  l  are -469.15 kJ/mol, -167.2 kJ/mol, -407.27 kJ/mol, and -285.83 kJ/mol, respectively.H =  Hf products  -  Hf reactants H = [ -407.27 kJ/mol  +  -285.83 kJ/mol ] - [ -469.15 kJ/mol  +  -167.2 kJ/mol ]H = -693.1 kJ/mol + 636.35 kJ/molH = -56.75 kJ/molNow, we can calculate the total enthalpy change for the reaction using the moles of the limiting reactant  HCl :Total enthalpy change = H  moles of limiting reactantTotal enthalpy change = -56.75 kJ/mol  0.050 mol = -2.8375 kJTo find the maximum temperature change, we need to calculate the heat capacity of the solution. Assuming the heat capacity of the solution is similar to that of water  4.18 J/gC , we can calculate the heat capacity using the mass of the solution. The mass of the solution is approximately the sum of the mass of NaOH and the mass of HCl solution:Mass of HCl solution = volume  density  50 mL  1 g/mL = 50 gMass of solution = 25 g + 50 g = 75 gNow, we can calculate the maximum temperature change using the total enthalpy change and the heat capacity of the solution:q = mcTT = q /  mc T = -2.8375 kJ /  75 g  4.18 J/gC T = -2837.5 J /  313.5 J/C T  -9.05CThe maximum temperature change the student can expect to see is approximately -9.05C, and the enthalpy change of the reaction is -56.75 kJ/mol.