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Question

r7 che XM l Id Base Titration Curves. · : Ac -Blinn College, Bryan Carpi x em × ng.com/ibisc ms/mod/ibis/view.php?id=5063535 0 2/28/2018 01:25 PM A 7.7/10 2/27/2018 08:50 PM Print Calculator-Penodic Table Question 4 of 13 ncorrect Sapling Learning At 25 C, the equilibrium partial pressures for the following reaction were found to PA-527 bar, PS = 5.80 bar, PC = 4.88 bar, and Pb = 4.58 bar. What is the standard change in Gibbs free energy of this reaction at 25 C? Number k.J/ mol rxn There is a hint available! View the hint by clicking on the bottom divider bar Click on the divider bar again to hide the hin.

Explanation / Answer

Ans. #Step 1: Calculate all partial pressures in terms of atm.

            PA = 5.27 bar = (5.27 / 0.986923) atm = 5.201 atm

            pB = 5.80 bar = (5.80 / 0.986923) atm = 5.724 atm

            PC = 4.88 bar = (4.88 / 0.986923) atm = 4.816 atm

            PD = 4.58 bar = (5.27 / 0.986923) atm = 4.520 atm

# Step 2: Calculate equilibrium constant, Kp

Equilibrium constant, Kp = (PC)4 (PD) / [PA (PB)2]

            Or, Kp = (4.816)4 x 4.520 / [5.201 x (5.724)2] = 14.269

# Step 3: Determine dG0

Using the equation dG0 = - RT lnKeq                  - equation 1

Where, dG0 = standard/ theoretical free energy change

T = temperature in kelvin = (0C + 273.15) K

Keq = Kp = equilibrium constant under standard condition

R = (0.001987 kcal mol-1K-1 or 0.008315 kJ mol-1 K-1)

Putting the values in equation 1-

            dG0 = - (0.008315 kJ mol-1 K-1) x 298.15 K x ln 14.269

            Or, dG0 = - (0.008315 kJ mol-1 K-1) x 298.15 K x (2.6581)

            Hence, dG0 = -6.59 kJ mol-1

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