A reaction $R I P$ has relative energies (kJ/mol): R = 0, TS1 = 72, I = 30, TS2 = 55, P = $- 40$ . Taking the effective activation energy of the overall reverse reaction ( $P R$ ) as the energy of its highest transition state relative to P, this is

A95 $kJ mo l^{- 1}$
B42 $kJ mo l^{- 1}$
C112 $kJ mo l^{- 1}$
D72 $kJ mo l^{- 1}$

Answer: C

The highest transition state on the path is TS1 at 72 kJ/mol, and the lowest preceding state of the reverse path is P at -40 kJ/mol. The effective barrier is 72-(-40) = 112 kJ/mol. The value 95 is only the first reverse step P to I, and 42 is the barrier from I.

Practice General Organic Chemistry III: Reaction Intermediates More on Reaction energy profiles, kinetic vs thermodynamic control

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A reaction R​I​P has relative energies (kJ/mol): R = 0, TS1 = 72, I = 30, TS2 = 55, P = −40. Taking the effective activation energy of the overall reverse reaction (P​R) as the energy of its highest transition state relative to P, this is

Answer: C

More questions on Reaction energy profiles, kinetic vs thermodynamic control

A reaction $R I P$ has relative energies (kJ/mol): R = 0, TS1 = 72, I = 30, TS2 = 55, P = $- 40$ . Taking the effective activation energy of the overall reverse reaction ( $P R$ ) as the energy of its highest transition state relative to P, this is