Equilibrium
In aquaculture tanks where excess ammonia is toxic, to keep water quality high without frequent changes, would an increase or decrease in pH shift the equilibrium to remove ammonia more effectively?
Adding more water to dilute the tank which may temporarily lower concentrations but doesn’t directly affect chemical equilibrium of ammonia.
Decreasing the pH to shift the equilibrium towards the ionized form of ammonia which is less volatile and toxic.
Increasing pH hoping to volatilize excess ammonia gases but ignoring impact on fish health or dissolved oxygen levels.
Reducing temperature to lower the molecular motion but without significant implications on the equilibrium between ammonium ions and anhydrous ammonia.
In an equilibrium expression, what does the square brackets around a chemical species signify?
[Number of moles]
[Concentration]
[Pressure]
[Volume]
How does the addition of a catalyst affect the position of equilibrium and value of equilibrium constant (K) for a given reversible reaction at constant temperature?
It increases the value of K while not changing the equilibrium position because the reaction proceeds more rapidly.
It does not cause any shift in equilibrium position nor change in K since it only increases the rate of reaching equilibrium.
It has no effects on the rate of reaching equilibrium but leads to a lower value of K due to increased concentrations of products.
It decreases the value of K while not changing the equilibrium position because it allows the reverse reaction to proceed more rapidly.
When determining the molecular shape of a compound with one central atom, two bonding pairs, and two lone pairs, what is the predicted geometry around the central atom?
Linear
Bent
Trigonal planar
Trigonal bipyramidal
In an endothermic reversible reaction, what happens to K when temperature is increased?
There is no change in the value of K because temperature doesn't influence equilibria constants directly.
The value of K becomes zero as all reactants are converted into products with sufficient thermal energy inputted into system.
The value of K decreases since increased kinetics favors reactant formation over product formation.
The value of K increases as more product forms at higher temperatures due to Le Chatelier's Principle.
If a reaction has a K value of 10 and the reaction quotient Q is calculated to be 20, what can be said about the concentrations of reactants and products?
The concentration of reactants is greater than the concentration of products.
The concentrations of reactants and products are equal.
It is impossible to determine the concentrations of reactants and products from this information.
The concentration of products is greater than the concentration of reactants.
How does adding a catalyst to a reversible reaction at equilibrium affect both forward and reverse reaction rates as well as the value of K?
Rates increase equally; K remains unchanged
Forward rate increases only; K increases
Both rates increase but not equally; K remains unchanged
Reverse rate increases only; K decreases

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What effect would decreasing the concentration of one of the reactants have on the reaction quotient Q for a reversible reaction currently not at equilibrium?
No effect on Q, but adjusts rates until equilibrium is established again
Increases Q without changing the direction of the shift until Q surpasses K
Shifts away from products until Q = K
Decreases until Q matches K, without any directional changes
What does an equilibrium constant (K) value greater than one indicate about a system at equilibrium?
[The concentration of reactants equals that of products.]
[The concentration of products is greater than that of reactants.]
[There are no reactants present.]
[The concentration of reactants is greater than that of products.]
A reaction has an equilibrium constant K of 2.5 x 10^-3 at a certain temperature. If the reaction quotient Q is calculated to be 8.0 x 10^-4, what can be said about the direction of the reaction?
The reaction is product-favored
The reaction is at equilibrium
The direction of the reaction cannot be determined from this information
The reaction is reactant-favored