Question

Calculate the entropy change (∆S) for the reaction of 2 moles of solid NH4Cl with 1 mole of NaOH in a coffee cup calorimeter at a constant temperature of 25°C. The reaction takes place in 100 mL of water and produces 2 moles of NH3 and 1 mole of NaCl in solution. Assume that the specific heat capacity of water is 4.18 J/g·K and that the final solution temperature after the reaction is complete is 25°C.

Answer 1

To calculate the entropy change  S  for the reaction, we need to consider the entropy change of the system  S_system  and the entropy change of the surroundings  S_surroundings . Since the final temperature of the solution remains constant at 25C, the entropy change of the surroundings will be zero. Therefore, we only need to calculate the entropy change of the system.The reaction is as follows:2 NH4Cl  s  + NaOH  s   2 NH3  aq  + NaCl  aq  + H2O  l First, we need to find the standard molar entropies  S  for each substance involved in the reaction. The standard molar entropies can be found in a standard thermodynamic table. The values are as follows:S  NH4Cl, s  = 94.6 J/molKS  NaOH, s  = 64.5 J/molKS  NH3, aq  = 80.8 J/molKS  NaCl, aq  = 115.5 J/molKS  H2O, l  = 69.9 J/molKNow we can calculate the entropy change of the system  S_system  using the following equation:S_system =  S products  -  S reactants S_system = [2  S NH3, aq  + S NaCl, aq  + S H2O, l ] - [2  S NH4Cl, s  + S NaOH, s ]S_system = [2  80.8 + 115.5 + 69.9] - [2  94.6 + 64.5]S_system = [277.1] - [253.7]S_system = 23.4 J/KSince the entropy change of the surroundings is zero, the total entropy change  S  for the reaction is equal to the entropy change of the system:S = S_system = 23.4 J/K