## 7.11 HW help

$K = \frac{k_{forward}}{k_{reverse}}$

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

### 7.11 HW help

(b) Sketch a reaction profile for the overall reaction, which is known to be exothermic. Label the activation energies of each step and the overall reaction enthalpy.

Basically, in the reaction mechanism, there are three steps total: (1) fast, (2) slow (RDS), (3) fast.

Does the "steepness" of the curve's drop in each step matter (aka whether it's endothermic or exothermic), as long as the overall reaction is exothermic? For instance, at first, I drew the second step as being extremely exothermic (a steep curve dropping), with the third step being endothermic (a slight curve with the products having more potential energy). However, in the answer key, the third step is the one where the curve drops steeply off to form an exothermic reaction.

Would the last step in the mechanism (3) have to be the one that exhibits an exothermic nature, since the products need to have lower potential energy than the reactants? Wouldn't there be no way to tell unless you're actually given the standard enthalpy values of each individual elementary reaction?

Alex Tchekanov Dis 2k
Posts: 118
Joined: Sat Aug 24, 2019 12:16 am

### Re: 7.11 HW help

As long as the initial reactants is higher than the final products, the heights of the intermediates do not matter

805303639
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### Re: 7.11 HW help

The reaction profile displayed in the solution manual represents a more likely scenario than the one you described. The third step will not be endothermic (products having more potential energy) because doing so necessitates that an intermediate (reactant in step 3) be more thermodynamically stable than the reaction's product.

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