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Equilibrium – Where? vs. Why?

Equilibrium – Where? vs. Why?. The value of K is an indication of WHERE to equilibrium rests. We haven’t addressed WHY the equilibrium exists where it does. Δ G. A ball rolling down two different hills. Potential energy “drives” the ball to the point of minimum potential energy.

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Equilibrium – Where? vs. Why?

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  1. Equilibrium – Where? vs. Why? The value of K is an indication of WHERE to equilibrium rests. We haven’t addressed WHY the equilibrium exists where it does. ΔG

  2. A ball rolling down two different hills. Potential energy “drives” the ball to the point of minimum potential energy.

  3. A ball rolling down two different hills. In the same way, free energy (ΔG) of a chemical reaction decreases until it reaches a minimum value. “Chemical Potential”

  4. The ball will come to rest at the position of lowest potential energy. This drawing is called a free-energy curve. A reaction will proceed to the point where free energy is lowest.

  5. DG vs DGo • DG applies only when the reactants and products are in their standard states. • Their normal state at that temperature • Partial pressure of gas = 1 atm • Concentration = 1 M • DG calculated for given (current) conditions. • Only one value for DG for a given reaction. • Many possible values for DG.

  6. N2O4 2NO2

  7. ΔG ΔG = ΔG + RT lnQ • R = 8.31 J/mol.K

  8. Consider the reaction at 298 K: 2H2S(g) + SO2(g) 3S(s, rhombic) + 2H2O(g) DGorxn = -102 kJ Calculate DGrxn under these conditions: PH2S = 2.00 atm PSO2 = 1.50 atm PH2O = 0.100 atm

  9. ΔG ΔG = ΔG + RT ln Q At equilibrium, ΔG = 0 ΔG = -RT ln K • and Q = K

  10. Calculate Kp for the reaction 2H2O(l) 2H2(g) + O2(g) Given that ΔG[H2O(l)] = -237.1 kJ/mol

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