6. Equilibrium DONE Flashcards
What is the end goal of reversible reactions (in terms of energy and entropy)?
Reversible reactions eventually reach a state in which energy is minimized and entropy is maximized and Gibbs free energy of the system is at a minimum.
What is entropy?
The measure of distribution of energy throughout a system or between a system and its environment
What is the difference between dynamic and static equilibrium?
Dynamic Equilibrium: Chemical reactions are going both forward and backwards
Static Equilibrium: The chemical reactions stop at equilibrium.
What is the law of mass action?
the principle that the rate of a chemical reaction is proportional to the masses of the reacting substances.
In terms of the law of Mass Action, what’s the difference between Q and Keq?
Q: A calculated value that relates the reactant and product concentrations at any given time during a reaction
Equilibrium Constant (Keq) is the ratio of products to reactants at equilibrium, with each species raised to its stoichiometric coefficient. Keq for a reaction is constant at a constant temperature
In order to calculate this, you need the equation and the number of moles…
How do you calculate the Equilibrium constant? Keq
[C,D,A,B] are the given concentrations, moles/L, while c,d,a,b (exponents) are the moles
***This is NOT the rate of the reaction, where we need to find the rate empirically. Instead, we can find equilibrium by this formula using the moles to guide us.
What is the reaction quotient? What’s the difference of this when compared to the equilibrium constant?
This can be done at any time during the reaction
, and looks exactly like the last equation.
The difference is that the equilibrium constant is only calculated at equilibrium, while the equilibrium constant, Qc, can be calculated at any time of the reaction
If you compare the reaction quotient and the equilibrium state, what will it tell you?
What will Q < Keq tell you?
What will Q = Keq tell you?
What will Q > Keq tell you?
It will tell you whether it will move back or forth.
If Q < Keq, ∆G < 0, and the reaction proceeds in the forward reaction
This makes sense. Because, when calculating Q, the numerator is the products, and the denominator is the reactants. If you have more reactants, the denominator is larger, and Q is smaller, and Keq is larger.
If Q = Keq, ∆G = 0, and the reaction is in dynamic equilibrium
If Q > Keq, ∆G > 0, and the reaction proceeds in the reverse direction, to create more reactants.
What will the (Keq) tell you?
If Keq > 1, the products are present in greater concentration at equilibrium
If Keq = 1, products and reactants are both present at equilibrium at reasonably similar levels
If Keq < 1, the reactants are present in greater concentration at equilibrium
If Keq ««_space;1 , the amount of reactants that have been converted to products can be considered negligible in comparison to the initial concentration of reactants.
What is the equilibrium constant of pure solids and liquids?
it is 1
So if you have a pure solid or liquid, you do NOT add it to the equation. You can, but it will not make a difference (because it is 1).
True or false: (Keq) is dependent on the temperature, meaning it will change with the temperature
True!
What would a very large (Keq) tell you? Such as one with a very large exponent?
Similarly, what would a very large (Keq) NEGATIVE exponent tell you?
If a K value has a large positive exponent, it’s likely that it has almost ran to completion.
If a K value has a large negative exponent, it’s likely that it’s retaining the majority of its reactants
Think of the ratio of (Keq)= products/reactants
Consider the hypothetical reaction A↔ B + C. for each of the following, determine if the amount of reactant A that has converted to product at equilibrium will be negligible compared to the starting concentration of A
The concentration of a reactant that converts to product can be considered negligible if it is two or more orders of magnitude less than the initial concentration of the reactant.
Concept Check 6.1.1: Given that [product] = .075 M and [reactant] = 1.5 M, determine the direction of reaction and the sign of the free energy change for reactions with the following Keq values: (Note: Assume that the reaction only has only one product and one reactant, and that the stoichiometric coefficient for each is one)
remember, Q= .05 (0.75/1.5)
Concept Check 6.1.2: Write the equilibrium constant expression for the following Reactions
What are four ways that this reaction will shift to the left or to the right…
- addition of reactants or products
- Removal of reactants or products
- increase or decrease pressure
- increase or decrease temp
Concept Check: 6.2.1 Describe what would happen in the following situations?
- Shift to the right, you are decreasing the free H+ ions, so would shift right
- Shift to the right, you are decreasing the pressure, so it will shift to the side with more moles
- Shift to the left, this is a combustion reaction, and heat is created, so if you create heat, it will go the the other way
- Will shift left, you are removing the reactants, so will shift to the left.
When thinking about kinetic products and thermodynamic control (think of the graph):
What will help you get to the kinetic product?
Answer: short times or low temperatures
Background:
Kinetic Products Definition: - higher in free energy than thermodynamic products and can form at lower temperatures. These are sometimes termed ‘fast’ products because they can form more quickly under such conditions.
When thinking about kinetic products and thermodynamic control (think of the graph):
What will help you get to the thermodynamic Product?
Answer: long time or high temperature
Background:
Thermodynamic products are lower in free energy than kinetic products and are therefore more stable. Despite proceeding more slowly than the kinetic pathway, the thermodynamic pathway is more spontaneous (more negative ∆G)
Over time, all of the kinetic products will go towards the thermodynamic products, which is why a long time will favor these products
When thinking about kinetic products and thermodynamic control (think of the graph):
What are Kinetic Products?
Higher in free energy than thermodynamic products and can form at lower temperatures. These are sometimes termed ‘fast’ products because they can form more quickly under such conditions.
When thinking about kinetic products and thermodynamic control (think of the graph):
What are Thermodynamic Products?
Are lower in free energy than kinetic products and are therefore more stable. Despite proceeding more slowly than the kinetic pathway, the thermodynamic pathway is more spontaneous (more negative ∆G)
Concept Check 6.3.2:
On a reaction coordinate diagram, how would the kinetic pathway appear as compared to the thermodynamic pathway?
Think about this in terms of activation energy and Gibbs Free Energy
The kinetic pathway would have a smaller activation energy, but also a small Gibbs free energy when compared to the thermodynamic product.
Book answer: Kinetic pathways require a smaller gain in free energy to reach a transition state. They also have a higher free energy of the products, with a smaller difference in free energy between the transition state and the products.
