- The equilibrium constant for the reaction is extremely large:
∆G° = –2.303 RT log Keq ⇒ log Keq = 2.303GRT
−∆ o
log Keq = 2.303G RT
−∆ o= 2.303x 0.00831(^100 kJ K^ kJ -mol (^1) mol)- (^1) x 333 K
− − -1
= 15.7Keq = 5.0 × 10^15
R = 0.08206 L atm mol–1 K–1 = 8.3143 j mol–1 K–11) A reaction goes to completion with such a large equilibrium constant.
2) The energy of the reaction goes downhill.- If covalent bonds are broken in a reaction, the reactants must go up an energy
hill first, before they can go downhill.
- A free-energy diagram: a plotting of the free energy of the reacting particles
against the reaction coordinate.
Figure 6.1 A free-energy diagram for a hypothetical SN2 reaction that takes place
with a negative ∆G°.
2) The reaction coordinate measures the progress of the reaction. It represents the