Question

true or false? exothermic reactions are always exergonic.
○ true
○ false

question 3
20 pts
which one of the following actions does not lead to an increase of the reaction rate (does not speed up the reaction)?
○ increasing the temperature of the reaction mixture.
○ using a catalyst.
○ increasing the concentration of the reactants.
○ using larger (coarser) pieces of a solid reagent.

Explanation:

First Question (True or False: Exothermic reactions are always exergonic)

Exothermic reactions release heat ($\Delta H < 0$), while exergonic reactions have negative Gibbs free energy change ($\Delta G < 0$). $\Delta G=\Delta H - T\Delta S$. If a reaction is exothermic ($\Delta H < 0$) but has a large negative $\Delta S$ (decrease in disorder) and high temperature, $T\Delta S$ could be positive enough to make $\Delta G>0$ (endergonic). However, in most cases at standard conditions, exothermic reactions tend to be exergonic, but the statement "always" is incorrect? Wait, no—wait, actually, exothermic means $\Delta H < 0$, and exergonic is $\Delta G < 0$. The formula $\Delta G=\Delta H - T\Delta S$. For exothermic ($\Delta H < 0$), if $\Delta S$ is positive, then $\Delta G$ is more negative (exergonic). If $\Delta S$ is negative, then at low $T$, $T\Delta S$ is small (since $T$ is low), so $\Delta G=\text{negative} - (\text{low }T)(\text{negative})=\text{negative} + (\text{low }T\text{ positive})$. So if $|\Delta H| > T|\Delta S|$, then $\Delta G < 0$ (exergonic). But is there a case where exothermic is not exergonic? Let's think of a reaction where $\Delta H=-10\ \text{kJ/mol}$, $\Delta S=-100\ \text{J/(mol·K)}$, and $T = 200\ \text{K}$. Then $T\Delta S=-20\ \text{kJ/mol}$. So $\Delta G=\Delta H - T\Delta S=-10 - (-20)=+10\ \text{kJ/mol}$ (endergonic). Wait, but $\Delta S$ here is negative (decrease in disorder). So in this case, exothermic but endergonic. Wait, but is this a real reaction? Maybe, but the key is the statement "always"—so exothermic reactions are NOT always exergonic. Wait, but I think I made a mistake earlier. Wait, no—let's check the definitions. Exothermic: heat released ($\Delta H < 0$). Exergonic: free energy released ($\Delta G < 0$). The relationship is $\Delta G=\Delta H - T\Delta S$. So if $\Delta H < 0$ and $\Delta S < 0$, then at high $T$, $T\Delta S$ (which is negative times positive $T$ is negative, so $-T\Delta S$ is positive) could make $\Delta G$ positive. For example, $\Delta H=-5\ \text{kJ/mol}$, $\Delta S=-20\ \text{J/(mol·K)}$, $T = 300\ \text{K}$. Then $T\Delta S=-6\ \text{kJ/mol}$, so $\Delta G=\Delta H - T\Delta S=-5 - (-6)=+1\ \text{kJ/mol}$ (endergonic). So exothermic (releases heat) but endergonic (requires free energy, non - spontaneous). Therefore, the statement "Exothermic reactions are always exergonic" is False.

• Increasing temperature: Increases kinetic energy of molecules, more frequent and energetic collisions, so reaction rate increases.
• Using a catalyst: Lowers activation energy, so more molecules have enough energy to react, rate increases.
• Increasing reactant concentration: Increases number of reactant molecules per unit volume, more frequent collisions, rate increases.
• Using larger (coarser) solid reagent: Decreases the surface area of the solid reactant. For reactions involving solids, reaction occurs at the surface. Larger pieces have less surface area, so fewer collisions at the surface, reaction rate decreases (does not speed up the reaction).

Answer:

False

Second Question (Which action does not speed up reaction)