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For the reaction N2O4(g) → 2 NO2(g), calculate delta-G (in J/mol) at a total pressure of 300 kPa and 27 deg C when gases are present in stoichiometric mole ratio (1 mol N2O4: 2 mol NO2). Given: delta-Gf(N2O4) = 100 kJ/mol, delta-Gf(NO2) = 50 kJ/mol, R = 8 J/(mol K), ln 2 = 0.7. Then compute the repeated digit-sum of this answer until a single digit remains.

1. 0
2. 1
3. 2
4. 3

Correct answer: 3

Solution

delta-G_std = 2*50 - 100 = 0 kJ/mol. At 300 kPa total, stoichiometric ratio 1:2: P_N2O4 = (1/3)*300 = 100 kPa; P_NO2 = (2/3)*300 = 200 kPa. Using P_std = 100 kPa: Q = (200/100)² / (100/100) = 4/1 = 4. delta-G = 0 + (8)(300)*ln(4) = 2400*2*ln(2) = 2400*2*0.7 = 3360 J/mol. Digit sum of 3360: 3+3+6+0 = 12; 1+2 = 3.

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