Question

Calculate the standard enthalpy change for the reaction where 50 ml of 0.1 M hydrochloric acid reacts with 50 ml of 0.1 M sodium hydroxide solution. The temperature change observed during the reaction was 5°C and the specific heat capacity of the solution is 4.18 J/g°C.

Answer 1

To calculate the standard enthalpy change for the reaction, we need to first find the heat released or absorbed during the reaction  q  and then divide it by the number of moles of the limiting reactant.1. Calculate the mass of the solution:The total volume of the solution is 50 ml  HCl  + 50 ml  NaOH  = 100 ml.Assuming the density of the solution is approximately equal to that of water  1 g/ml , the mass of the solution is 100 ml  1 g/ml = 100 g.2. Calculate the heat released or absorbed  q  during the reaction:q = mass  specific heat capacity  temperature changeq = 100 g  4.18 J/gC  5Cq = 2090 J3. Determine the limiting reactant and calculate the number of moles:Both HCl and NaOH have the same concentration  0.1 M  and volume  50 ml , so they will react in a 1:1 ratio. Therefore, there is no limiting reactant in this case.Number of moles of HCl = concentration  volume  in liters Number of moles of HCl = 0.1 M  50 ml   1 L / 1000 ml Number of moles of HCl = 0.005 mol4. Calculate the standard enthalpy change  H  for the reaction:H = q / number of molesH = 2090 J / 0.005 molH = 418000 J/molThe standard enthalpy change for the reaction is 418 kJ/mol.