18.4.1 Enthalpy and Entropy Contributions to K Flashcards
Enthalpy and Entropy Contributions to K
- The standard Gibbs free energy change of reaction (ΔG°) is related to the equilibrium constant (K eq ).
- The equilibrium constant is related to both the ΔH° and the ΔS° in the following way: ln K eq = ΔH°/(RT) + ΔS°/R
- At high temperatures, ΔS° determines whether ΔG° is positive or negative; at low temperatures, ΔH° determines whether ΔG° is positive or negative.
note
- The standard Gibbs free energy change of reaction (ΔG°) is related to the equilibrium constant (K eq ).
- When ΔG° = 0, the reaction is at equilibrium. When the system is at equilibrium, K eq = 1.
- When ΔG° > 0, the reaction will run toward reactants.
Therefore, K eq < 1. - When ΔG° < 0, the reaction will run toward products.
Therefore, K eq > 1. - The equilibrium constant (K eq ) is related to both ΔH° and ΔS°. The equilibrium constant depends directly on ΔS° and is modified by ΔH°.
- When T is very high, the ΔH°/RT term approaches 0 and the ΔS° term predominates in this equation. In such a situation, the equilibrium is based on where there is the largest entropy.
- Conversely, when T is very low, the ΔH°/RT term becomes very large and, since ΔS°/R has no temperature component, the enthalpy term will predominate in the equation as the reaction searches for the lowest chemical energy.
- When ΔS° < 0, entropy favors the reactants.
- An exothermic reaction with ΔS° < 0 favors reactants (ln K < 0) at high temperatures, but favors products (ln K > 0) at low temperatures.
- However, an endothermic reaction with ΔS° < 0 always favors reactants (ln K is always less than 0).
- When ΔS° > 0, entropy favors the products.
- An exothermic reaction with ΔS° < 0 always favors products (ln K is always greater than 0).
- An endothermic reaction with ΔS° > 0 favors products (ln > 0) at high temperatures, but favors reactants (ln K < 0) at low temperatures.
Which of the following is not correct?
ΔG° = −RT log 10 (Keq )
Look at the diagram for ln (Keq ) = −ΔH° / RT + ΔS° / R, in which ΔS° > 0.
Which statement about this diagram is not correct?
For the endothermic reaction, no matter what we do (with temperature), ΔH° and ΔS° always favor the products
Which statement about the plots for lnKeq = −ΔH° / RT + ΔS° / R, in which ΔS° < 0, is correct?
For the endothermic reaction, the reactants are always favored.
The expression ln (Keq ) = −ΔH° / RT + ΔS° / R relates the value of ΔG to the Keq term. Which statement about this expression for ln (Keq ) is not correct?
The value of ΔH° / RT increases with increasing temperature
Look at the snake analogy drawing.
Which statement about this analogy is not correct?
This diagram represents an exothermic reaction.
Which statement (or equation) about Gibbs free energy and related topics is not correct?
ln (Keq ) = ΔH° / R − ΔS° / RT
Look at the diagram that shows ln (KEq ) = −ΔH° / RT + ΔS° / R for an endothermic reaction and an exothermic reaction where ΔS° < 0.
Which statement about this reaction is not correct?
By changing the temperature in the endothermic reaction where ΔS° < 0, we can influence whether products or reactants are favored at equilibrium
Now look at a snake analogy in which there is a small, elevated table in the middle of the snake pit.
Which statement about this analogy is not correct?
If we increase the temperature enough, all of the snakes can be simultaneously located on the upper level.
Which statement about the expression ln (Keq ) = −ΔH° / RT + ΔS° / R is not correct?
At low temperatures, the equilibrium constant is based on maximized entropy.
Look at the diagram that shows ln (Keq ) = −ΔH° / RT + ΔS° / R for an endothermic reaction and an exothermic reaction where ΔS° > 0.
Which statement about this reaction is not correct?
For the exothermic reaction, the value of ln (Keq ) increases from right to left. For the endothermic reaction, the value of ln (Keq ) increases from left to right.
