Entropy help? ASAP please? due tomorrow?

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Calculate change of S for the reaction: 2NO+ O2 =2NO2 values for each in J/mol*K NO=240.0 O2=205.2 NO=210.8 The equation to solve the problem is as follows: Change ...show more
Update : Thank you all- now can you tell me what change of Sf is? I know that it ...show more
Update 2: Thank you so much Dr. Lee!
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n is the coefficient in front of the species in the balanced equation....

so you take the number of each product and multiply it by the standard entropy of formation (Sf) and add them all together

then take the number of each reactant and multiply it by the standard entropy of formation (Sf)

then subtract the reactant entropy from the product entropy

so for what you have up there:

delta S = (2(Sf NO2)) - (2(Sf NO) + 1(Sf O2))

where the numbers are the numbers in front of each species from the balanced equation you provided in the question

you have NO listed twice and i don't have a textbook to see which is NO and which is NO2 but that's how you set it up

good luck!!

Sf is the standard entropy of formation...when you make a molecule from it's component elements in their standard states, that is defined as "formation"....so like N2 (g) + O2 (g) --> 2 NO (g) since N2 and O2 are gases in their standard states at 25C and 1 atm of pressure...so when this happens there is a change in entropy since the products are different from the reactants...this change in entropy is the standard entropy of formation


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Other Answers (1)

  • RossK answered 5 years ago
    To find the entropy (delS) for the reaction you need to subtract the entropy of the reactants from the entropy of the products

    delS = 2 mol (NO2) X 210.8 - ((2 mol (NO) X 240.0 + 1 mol (O2) X 205.2)) = 421.6 - 480.0 - 205.2 = -263.4.

    That is if the entropy for the second "NO" (210.8) is actually entropy for NO2.

    Oh yes, n = number of moles.
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