Question

Calculate the standard enthalpy change for the reaction between 50.0 mL of 0.200 M HCl(aq) and 50.0 mL of 0.300 M NaOH(aq) at 25°C. The heat capacity of the calorimeter is 9.60 J/°C and the temperature change observed is 4.00°C. Also, assume that the density of both solutions is 1.00 g/mL and that the heat capacity and density of the final solution are the same as that of water.

Answer 1

To calculate the standard enthalpy change for the reaction, we first need to determine the amount of heat released during the reaction  q  and the number of moles of the limiting reactant that reacted.The balanced chemical equation for the reaction is:HCl aq  + NaOH aq   NaCl aq  + H2O l First, we need to find the limiting reactant. We can do this by calculating the moles of each reactant:moles of HCl = volume  L   concentration  M moles of HCl = 0.050 L  0.200 M = 0.0100 molmoles of NaOH = volume  L   concentration  M moles of NaOH = 0.050 L  0.300 M = 0.0150 molSince there are fewer moles of HCl, it is the limiting reactant.Next, we need to calculate the heat released during the reaction  q . We can do this using the heat capacity of the calorimeter and the temperature change:q = heat capacity  temperature changeq = 9.60 J/C  4.00C = 38.4 JNow, we can calculate the enthalpy change per mole of the limiting reactant  HCl :H = q / moles of limiting reactantH = 38.4 J / 0.0100 mol = 3840 J/molSince the reaction is exothermic  heat is released , the standard enthalpy change for the reaction is negative:H = -3840 J/molSo, the standard enthalpy change for the reaction between 50.0 mL of 0.200 M HCl aq  and 50.0 mL of 0.300 M NaOH aq  at 25C is -3840 J/mol.