Question

Calculate the standard entropy change for the following using the table of standard values.(first, predict the sign for ∆S qualitatively)

2H₂(g)+O₂(g)→2H₂O(g)

Answer 1

Answer -:

At 298K as a standard temperature:

• $S\degree$(H₂) = 130.68 J/mol K
• $S\degree$(O₂) = 205.14 J/mol K
• $S\degree$(H₂0) = 188.83 J/mol K

From the balanced equation we can write the equation for $S\degree$(the change in the standard molar entropy for the reaction):

$S\degree = 2*S\degree(H_2O) - [2*S\degree(H_2)+S\degree(O_2)]\\ S\degree=2*188.83 - [ 2*130.68+205.14]\\ S\degree= -88.84 J/mol K$

It would appear that the process result in a decrease in entropy -i.e. a decrease in disorder. This is expected because we are decreasing the number of gas molecules.