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View History Bookmarks Window Help saplinglearning.com Activities and Due Dates HW 19 iversity of Charlotte-CHM 112-Spring18-ACHESON 4/15/2018 11:00 PM 45.2/100 4/5/2018 11:19 AM Print l Calculator Periodic Table Question 15 of 18 Sapling Learning At 25 ?, the equilibrium partial pressures for the following reaction were found to be PA 4.04 atrn, Pe-5.17 atm, PC #5.55 atm, and Pb-4.90 atm. What is the standard change in Gibbs free energy of this reaction at 25 C? Number kJ/ mol Hint O Previous Q9 Give Up & View Solution 2, Check Answer 0 Next ?? Ex

Explanation / Answer

Solution :

We know that ,Gibbs free energy is given as:

delG = - RT lnQ

Here , T = 250C = 25+273.15=298.15 K

R= 0.008314 kJ / K

Q= (PC) (PD)2 /(PA)3 (PB)2

= (5.05)*(4.90)2/(4.04)3*(5.17)2

= 0.075606

Now, del G = - (0.008314)(298.15)ln(0.075606)

= 6.4008kJ

Therefore the Gibbs free energy is = 6.4008kJ/mol

  

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