Calculate the enthalpy change and the entropy change of the reaction where 2 moles of HCl(g) reacts with 1 mole of CaO(s) to produce 1 mole of CaCl2(s) and 1 mole of H2O(l). Given the bond dissociation enthalpies (in kJ/mol) of H-H is 436, Cl-Cl is 243, H-Cl is 431, and O-H is 463. The standard entropies (in J/mol K) of HCl(g) is 186.9, CaO(s) is 38.2, CaCl2(s) is 223.0, and H2O(l) is 69.9.

Question

Calculate the enthalpy change and the entropy change of the reaction where 2 moles of HCl(g) reacts with 1 mole of CaO(s) to produce 1 mole of CaCl2(s) and 1 mole of H2O(l). Given the bond dissociation enthalpies (in kJ/mol) of H-H is 436, Cl-Cl is 243, H-Cl is 431, and O-H is 463. The standard entropies (in J/mol K) of HCl(g) is 186.9, Ca

Answer 1

To calculate the enthalpy change for the reaction, we can use the bond dissociation enthalpies given. The reaction can be represented as:CaO s  + 2HCl g   CaCl2 s  + H2O l First, we need to determine the energy required to break the bonds in the reactants and the energy released when the bonds in the products are formed.For the reactants:1 mole of CaO has no bond dissociation enthalpy given, so we will assume it to be 0.2 moles of HCl: 2 * 431 kJ/mol = 862 kJTotal energy required to break bonds in reactants = 0 + 862 = 862 kJFor the products:1 mole of CaCl2 has no bond dissociation enthalpy given, so we will assume it to be 0.1 mole of H2O: 1 *  2 * 463  kJ/mol = 926 kJ  since there are 2 O-H bonds in H2O Total energy released when bonds in products are formed = 0 + 926 = 926 kJEnthalpy change  H  = Energy released by products - Energy required for reactantsH = 926 kJ - 862 kJ = 64 kJNow, let's calculate the entropy change for the reaction. We can use the standard entropies given.Entropy change  S  =  Entropy of products  -  Entropy of reactants For the reactants:1 mole of CaO: 38.2 J/mol K2 moles of HCl: 2 * 186.9 J/mol K = 373.8 J/mol KTotal entropy of reactants = 38.2 + 373.8 = 412.0 J/mol KFor the products:1 mole of CaCl2: 223.0 J/mol K1 mole of H2O: 69.9 J/mol KTotal entropy of products = 223.0 + 69.9 = 292.9 J/mol KS = 292.9 J/mol K - 412.0 J/mol K = -119.1 J/mol KSo, the enthalpy change  H  for the reaction is 64 kJ, and the entropy change  S  is -119.1 J/mol K.