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### Drawing and labeling reaction profile

Posted: Thu Feb 25, 2016 8:33 pm
4.35 says to draw and label the reaction profile for the reaction: CH3Br + -OH $\rightleftharpoons$ CH3OH + Br-.

For this particular reaction, how do we know that the Gibbs free energy of the reaction between the reactants and products is negative?

For other reactions, how will we know whether $\Delta$G will be positive or negative between the reactants and products?

### Re: Drawing and labeling reaction profile

Posted: Thu Feb 25, 2016 10:19 pm
$\Delta G_{r}$ is equal to $\Delta G_{products} - \Delta G_{reactants}.$

If the graph's final point is below the initial point, then that means that $\Delta G_{products} < \Delta G_{reactants}$ and therefore $\Delta G_{r}$ will be negative.

If the graph's final point is above the initial point, then that means that $\Delta G_{products} > \Delta G_{reactants}$ and therefore $\Delta G_{r}$ will be positive.

### Re: Drawing and labeling reaction profile

Posted: Sat Feb 27, 2016 3:00 pm
Alli Foreman 2H wrote:$\Delta G_{r}$ is equal to $\Delta G_{products} - \Delta G_{reactants}.$

If the graph's final point is below the initial point, then that means that $\Delta G_{products} < \Delta G_{reactants}$ and therefore $\Delta G_{r}$ will be negative.

If the graph's final point is above the initial point, then that means that $\Delta G_{products} > \Delta G_{reactants}$ and therefore $\Delta G_{r}$ will be positive.

How about for $\Delta$G$\circ$$\neq$

### Re: Drawing and labeling reaction profile

Posted: Sat Mar 12, 2016 12:03 am
$\Delta G\neq ^{\circ}$ is always positive. It is the difference between the $\Delta G$ of the reactants and the $\Delta G$ of the intermediates. $\Delta G$ intermediates is always above $\Delta G$ reactants (think of the graph, it's always a hill shape).
So $\Delta G$ of the intermediates - $\Delta G$ of the reactants is always positive.