Question

Calculate the entropy change of the following reaction at 25°C and 1 atm pressure:2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g)Given the standard entropy values are:ΔS°NH3 = 192.5 J/K•molΔS°Cl2 = 223.1 J/K•molΔS°N2 = 191.6 J/K•molΔS°HCl = 186.9 J/K•mol

Answer 1

To calculate the entropy change  S  for the reaction, we need to find the difference between the standard entropy values of the products and the reactants. The equation for this is:S reaction  = S products  - S reactants For the reactants, we have:2 moles of NH3: 2 * 192.5 J/Kmol = 385 J/Kmol3 moles of Cl2: 3 * 223.1 J/Kmol = 669.3 J/KmolFor the products, we have:1 mole of N2: 1 * 191.6 J/Kmol = 191.6 J/Kmol6 moles of HCl: 6 * 186.9 J/Kmol = 1121.4 J/KmolNow, we can calculate the entropy change for the reaction:S reaction  =  191.6 + 1121.4  -  385 + 669.3 S reaction  = 1313 - 1054.3S reaction  = 258.7 J/KmolThe entropy change for the reaction at 25C and 1 atm pressure is 258.7 J/Kmol.