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 -:
At 298K as a standard temperature:
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)]\\\\\nS\\degree=2*188.83 - [ 2*130.68+205.14]\\\\\nS\\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.
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