inais carbondale HEM 210- Fal16- WITH I Activities and Due Dates I workshop 1201

Question

inais carbondale HEM 210- Fal16- WITH I Activities and Due Dates I workshop 1201114 11/1a/2016 osoo PM G 22.21100 Calculator Periodic Table 5 of 9 Sapling Learning At 25 c, the equilibrium partial pressures for the following reaction were found to be PA 4.34 atm, Ps 5.37 atm, Pc 5.39 atm, and Po 4.26 atm. 3A(R) +2B(R) +20(R) What is the standard change in Gibbs free energy of this reaction at 25 "C? Number AG, kJ mol "AO Previous Check Answer 0Next a Er Would you like to save your password for saplngleamingoom? More info

Explanation / Answer

dGRXN = -RT*ln(Kc)

in equilibrium:

Kc = [C][D]^2 /[A]^3[B]^2

so:

Kp = P-C * P-D^2 / (P-A^3)(P-B)^2

Kp =5.39 * (4.26^2)/((4.34^3)(5.37^2)) = 0.041494

so..

Kp = Kc*(RT)^dn

dn = (1+2)-(3+2) = -2

solve for Kc

Kc = Kp*(RT)^2

Kc = (0.041494 )((0.082*298)^2)

Kc = 24.776

so..

dGRXN = -RT*ln(Kc)

DGrxn = -8.314*298*ln(24.776) =-7945.0859 J/mol = -7.945 kJ/mol

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