CH102 - Quiz 3
Quadratic Formula
-b±[√(b²-4ac)]/2a
Kc is never what?
0 and infinity!
Steps to Solve (ICE)
1. Container 2. Equation, Qc 3. Solve
Catalysts
1. Not consumed (they are there at the beginning and at the end). 2. They speed up reactions. 3. They do not lower activation energies! 4. They change the mechanism of the reaction and the new mechanism has lower activation energies.
For a reverse reaction, what does K new equal?
1/K
Endothermic (ChangeH > 0)
A + "heat" equil. B
Exothermic (ChangeH < 0)
A + B equil. C + "heat"
100% yield is what?
A limiting reagent problem. When doing this, start the steps over and do another problem using the same equation, Q, and Ksp.
At equilibrium, concentrations don't change, so that means Qc...
Also stops changing!
If Kc is small, then...
At equilibrium there are more reactants.
Small Kc
At equilibrium, few products, lots of reactants. Small % yield.
Large Kc
At equilibrium, lots of products, very few reactants. Large % yield.
If Kc is large, then...
At equilibrium, there are more products.
Equilibrium
At some point, the rate of the forward reaction is equal to the rate of the reverse reaction.
What do you need to make a judgement on spontaneity?
Both Qc and Kc.
Kc is a...
Constant!
All reactions get slower when you do what?
Cool them down!
If Qc is significantly bigger than K, it has to what?
Decrease to get to equilibrium.
At equilibrium, the concentration...
Does not change!
When adding water to a solution, Q goes...
Down because Qnew is smaller!
What is x?
How much product will be made or how big is the change to get to equilibrium.
What does Le Chatelier's Principle say about temperature?
If you increase the temperature, Kc changes and goes up.
What is the equivalent to increasing the volume of a gas-phase reaction for a reaction that occurs in solution?
Increasing the amount of solvent is equivalent to increasing the volume of a gas-phase reaction.
What does a catalyst do?
It fundamentally changes the mechanism of the reaction. The overall reaction is the same, but it goes from a single step reaction to a multi-step mecahnism.
The rate of the reverse reaction will not get bigger than the rate of the forward because...
It is already big! There is already a large percent of molecules that is able to overcome the barrier.
How can we use molar solubility?
It will give us x.
When adding solvents to a non-pure solution, will solubility increase or decrease?
It will lower because the concentration of products increases so we don't have to dissolve as much reactants for Qsp = Ksp. We are closer to equilibrium because there is already some products there.
If you multiply a reaction by x, what does Knew equal?
K^x
For reactions with only gases, we can use Kp instead of Kc.
Kc = Kp (1/RT)^change in the number of moles of gas in the reaction.
At equilibrium, Ratefwd = Raterev
Kfwd/krev = [B]/[A] = Kc.
What happens when you couple/add reactions and their equilibria (e.g. K1 + K2)?
Knew = K1 * K2
Equilibrium Pressure
Kp
For reactions with only gases, we can use what instead of Kc?
Kp.
Due to the catalyst, steps in the new mechanism have what?
Lower activation energies.
K2 > K1
More products and fewer reactants.
According to Le Chatelier's Principle, if you add more reactants concentration, what happens?
More products form because it makes Qnew smaller and Qnew < K. Spontaneous!
In an exothermic reaction, what happens when temperature decreases?
More products form!
According to Le Chatelier's Principle, if you add more product concentration, what happens?
More reactants form because it makes Qnew bigger and Qnew > K. Non-spontaneous!
According to Le Chatelier's Principle, if you remove reactants concentrations, what happens?
More reactants form because it makes Qnew bigger and Qnew > K. Non-spontaneous!
How much solute redissolves?
Multiply the liters by the number of ions produced (E) because this is the concentration. It will give you the moles dissolved. The number of moles in the beaker minus that number is how much is left.
How do you get the concentration of ions in a resulting solution?
Multiply the number of L of the solution times the concentration to get the number of moles and then divide that by the new volume of the combined solutions added.
Can a catalyst lower the enthalpy change for a reaction?
No because enthalpy is only related to the energy products and reactants. Catalysts only deal with activation energy.
Do all collisions with energies greater than the activation energy result in a reaction?
No because molecules also have to be in the correct orientation.
Non-Spontanteous
Products are consumed, reactants are being made.
In an endothermic reaction, what happens when temperature increases?
Products form!
Temperature does not change, what?
Q!
At the beginning of a reaction where there are only reactants ([products] = 0)...
Qc = 0.
At equilibrium...
Qc = Kc
If we only have products ([reactants] = 0)....
Qc is approaching infinity.
For any reaction that only has gases (or solids and pure liquids), we can write Qp in terms of what?
Qp!
What happens when you increase the volume of a container?
Qp, new is smaller because the pressure of gas product goes down.
Spontaneous
Reactants are going down, products are going up!
Kc is always...
Relative! Products compared to reactants.
Activation Energy
Stays the same because activation energy is a constant for a given reaction!
In dynamic chemical equilibriums, the reaction never what?
Stops!
Reaction Quotient (Qc)
Tells us the current state of the reaction (what is present in the beaker).
The reaction with the greatest activation energy is the most sensitive to what?
Temperature change.
What is the only thing that makes K change?
Temperature!
Some reactions have no activation energy, that is, Ea = 0. What would the rate of such a reaction depend on?
The Arrhenius factor or likelihood of collision.
What does molar solubility tell us?
The amount of moles of solute that dissolves in a liter of solvent.
The reaction never stops, but at equilibrium...
The concentration, rate of forward reaction, and rate of reverse reaction stop changing.
The number of moles don't change inside the beaker, but what does change?
The concentration.
If Q = K, then we are at equilibrium.
The concentrations and pressures do not change.
If Q =/= K, then change will occur to reach equilibrium.
The concentrations and pressures will change until Q = K.
Kc is not the equilibrium quotient!
The equilibrium quotient is what the reaction is right now.
If reactants are forming, that means what reaction gets slower?
The forward reaction because it has bigger activation energy. This means it must be endothermic.
When you open the valve and the volume of the container increases, what happens to water vapor?
The pressure of the water vapor decreases. Qnew is now less than Kp, and the reaction is spontaneous.
Changing the temperature changes K and not Q.
The rate of all reactions go up.
If you only have reactants...
The rate of the forward reaction is big and then decreases. The rate of the revers reaction is low and then increases over time.
Qc < Kc
The ratio of products over reactants now is lower than it will be at equilibrium (Kc). More products will be produced. Reactants are consumed!
Qc = Kc
The ratio of products over reactants now is not going to change anymore. More products will be produced. Reactants are consumed! Concentrations will not change anymore.
Qc > Kc
The ratio of products to reactants now is higher than what it will be when at equilibrium (Kc). The concentration of products will increase, the concentration of reactants will decrease, Qc will decrease.Non-spontaneous (revers reaction is now spontaneous).
If there's the same amount of gases in the products and reactants, then what happens when you change the volume?
The ratio stays the same.
Will the forward reaction or reverse reaction speed up more?
The reaction that has fewer molecules that have enough energy to get over the barrier. The rate constant that is smaller will go faster.
If products are forming, that means what reaction gets slower?
The reverse reaction because it has bigger activation energy. This means it must be exothermic.
Kc
The value of Qc at equilibrium.
If Qc is large, then...
There are currently more products.
If Qc is small, then...
There are currently more reactants.
When adding ions that form a precipitate with aqueous products, Qnew goes down!
There becomes less products formed because of precipitates.
What are some misconceptions of catalysts?
They do not increase the yield of a reaction, they just increase the rate in which the product forms. Catalysts do not lower the overall activation energy but provide new mechanisms in which have lower activation energy steps. A catalyst does participate in a reaction, but it does not show up in the overall reaction.
Solids do not show up in Qc or Kc because...
They don't have concentrations!
Heat is not a reactant!
This is a heuristic.
The reverse reaction gets more slow than the forward reaction.
This means Kc gets bigger.
If K is tiny, there are few products at equilibrium.
This means the change (x) is really small.
When the volume of a gas-phase reaction is increased (at constant pressure), in what direction does the equilibrium shift?
To the side with more moles of gas.
In an exothermic reaction, the activation energy of the reverse reaction is what?
Very large and the activation energy of the forward reaction is much smaller.
In an endothermic reaction, the activation energy of the forward reaction is what?
Very large and the activation energy of the reverse reaction is much smaller.
When K is huge and Q is less than K, what can we assume?
We can assume 100% yield.
We can assume the change is tiny.
We can then get rid of the initial concentration plus the x change and then solve for a single x.
Under what conditions does increasing the volume of a gas-phase reaction have no effect on the equilibrium state?
When moles of gas are equal on both sides.
If it tells you the molar solubility solve the ice table for Ksp and that does not change even in the new problem.
You can figure out Ksp from the molar solubility in first pure water.
P = [gas]RT
[gas] = Pgas/RT
Qc Equation
[products]/[reactants]
Kc Formula
kfwd/krev
Formula for Ea
ln(k2/k1) = -Ea/R(1/T2 - 1/T1)