Question

6) Some unimolecular reaction, A → B, has a AH = -31.2 kJ/mol and a AS = 4.8 J/mol K. At 25 °C the reaction has a rate constant of 0.155 5-1. Raising the temperature to 37 °C increases the rate constant to 0.216 s-. Draw the potential energy surface (in terms of AG vs reaction coordinate) for this reaction at 25 °C (assume Hand S do not change with temperature).

Answer 1

Here, $\Delta$H is negative and $\Delta$S is positive, hence the reaction is spontaneous (i.e. $\Delta$G will be negative) at any temperature.

According to the Gibbs-Helmholtz equation:

$\Delta$G = $\Delta$H - T.$\Delta$S

i.e. $\Delta$G = -31.2 kJ/mol - (25+273) K * 4.8 J/mol.K

i.e. $\Delta$G = -31.2 kJ/mol - 1.4304 kJ/mol

i.e. $\Delta$G = -32.6304 kJ/mol

Now, the potential for this reaction can be drawn as follows.

Activation energy AG Reactants (A) Products (B) Reaction Co-ordinates