CHEM211 Problem Set
Reaction Energetics
1) For the molecules below:
a) Draw the most stable and least stable conformations.
b) Calculate the difference in energy between the conformations.
c) Calculate the equilibrium constant (Keq) for conversion between the conformations at room
temperature (70 oF, 21 oC, 294 K).
(H3C)2HC CH3H3CH2C CH(CH3)2
t-Bu
OH
Cl
2) Sketch an energy diagram to match each of the following descriptions:
a) Fast reaction; ∆Go = -120 kJ/mol.
b) Slow reaction; ∆Go = 20 kJ/mol.
c) Very slow reaction; Keq = 1.36×10 –3.
d) Fast reaction; Keq = 3.0.
e) A two step reaction; first step is fast; second step is slow; ∆Go = -80 kJ/mol.
f) ∆G1‡ = 200 kJ/mol; ∆G2‡ = 50 kJ/mol; ∆Go = -40 kJ/mol.
3) For each energy diagram in (2), label starting material(s) (SM), product(s) (P), intermediate(s) (I),
transition state(s) (TS‡), activation energy(s) (∆
∆∆
∆G‡) and free energy (∆
∆∆
∆Go).
4) For the reactions below, calculate ∆Go, ∆Ho and ∆So.
Br Br
Br2
290 K Keq = 1.24 x 10 19
HBr
Br Keq = 3.96 x 10 15
Br OOH
Br Keq = 0.13