Card 0 of 866
Consider the following equation at equilibrium:
What would happen if more were added into the equation?
Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.
Before the addition of , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant;
.
However, after the addition of , a reactant, the reaction quotient is now less than the equilibrium constant;
.
In order to maintain equilibrium, the reaction will then shift right of favor the formation of more products.
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Consider the following equation at equilibrium:
What would happen if more were added to the system?
Recall Le Chatelier's principle: A chemical system at equilibrium will shift in the direction that minimizes the disturbance to the system.
Before the addition of , the system is in equilibrium, meaning the reaction quotient is equal to the equilibrium constant;
.
However, after the addition of , a product, the reaction quotient is now less than the equilibrium constant;
.
In order to maintain equilibrium, the reaction will then shift left to favor the reactants in this chemical system.
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Consider the following exothermic equation:
Which of the following actions will cause the products to be favored?
Since the equation is exothermic, you can think of heat as another product of the reaction:
Apply LeChatelier's principle to this equation.
If the temperature at which the reaction is decreased, that is akin to decreasing the amount of product made, thus causing the reaction to shift towards the products.
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What is the difference between the reaction quotient and the equilibrium constant?
The correct answer gives the accurate definition of both the equilibrium constant and the reaction quotient.
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Which of the following can always be determined based on the equation itself?
According to the law of mass action, the equilibrium constant expression can always be written given the equation of the reaction itself. The equilibrium-constant expression will be written as the products over the reactants, each raised to their respectively stoichiometric coefficient. The rate law and concentrations can only be determined if there is additional data given.
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The law of mass action gives an expression that is specific for a certain __________.
Equilibrium expressions are specific for certain temperatures. Physical state, pressure, and volume do not factor into the expression.
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Consider formation of Nitrogen monoxide: with
. If the initial concentration of N2 was 0.085M and O2 was 0.038 M, what is the concentration of nitrogen monoxide at equilibrium?
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Given the ICE table below, what are the signs on the terms that would appear in the “Change” row?
Based on the balanced equation, one would use -x, -2x, and +x.
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Find the equilibrium concentrations for C in the following chemical reaction: A + B -> 2C
K = 9.0 x 10-8
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What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.60 M?
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What is the equilibrium concentration for A- in the following reaction if the starting concentration of HA is 0.30 M?
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Consider the following gasesous, reversible, exothermic reaction:
What could be done to increase the equilibrium concentration of species C?
According to Le Chatelier's principle, any changes in concentration or pressure to a system at equilibrium will cause the system to readjust to a new equilbrium. The addition of a reactant will cause the reaction to shift to the right, increasing the equilibirum concentration of the products. Thus, adding additional reactant A would increase the equlibrium concentration of product C.
Removing species A or adding species E will drive the reaction to the left, reducing the amount of species C. Since the reaction is exothermic, heat can be considered a product. Thus, increasing the temperature will also shift the reaction to the left.
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The reaction is at equilibrium. You have measured the concentration of
and
to be
What is the equilibrium constant, ?
Recall that the rate constant for the equilibrium reaction
is , always remembering that it is products over reactants.
Since we are at equilibrium, for every mol of we have, we have 3 mols of
. Therefore,
Likewise, for every 2 mols of we have, we have 2 mols of
. Therefore,
Plugging all of this into the above equation we get
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Consider the following chemical reaction.
This reaction is allowed to equilibrate at some volume and pressure.
Which of the following actions will not shift the equilibrium toward the products?
Adding an inert gas will not shift the equilibrium toward the products, but neither will it shift the equilibrium toward the reactants. It has no effect because the partial pressure of each species, ,
, and
, will still remain constant. Since the partial pressures are constant, we know from the ideal gas law that the concentration of each species remains constant. For example, for
,
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A reaction at equilibrium releases 25 Joules of heat energy at . What is the entropy change for this same reaction?
To solve this problem, we must make use of the following equation:
At equilibrium, the value of is 0. Therefore, we can simplify the equation.
We must also convert temperature from degress Celsius into Kelvin.
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If formate is added to a solution of formic acid in water, which of the following changes would occur?
Because formic acid is a weak acid, it will exist in equilibrium with its conjugate base, formate, while in water according to the following reaction:
If formate is added to this solution, then according to Le Chatlier's principle, the reaction will shift towards the left. Thus, the in solution will decrease and the pH will consequently increase. Furthermore, the addition of formate has no effect on the pKa.
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For a given chemical reaction, . What is the
for this reaction at
?
To answer this question, it is necessary to use following equation:
We can solve for by rearranging the equation to the following:
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Which of the following is the correct equilibrium expression for the given reaction?
For this question, we're asked to determine the correct equilibrium expression for a given reaction that is in equilibrium.
To begin with, we need to remember that pure solids and liquids are omitted from the equilibrium expression. Only gases and aqueous components of a reaction are included. The reason for this is because in a reaction at equilibrium, a pure solid or liquid will not have any appreciable change in concentration. Because their concentrations remain essentially unchanged, their value gets incorporated into the constant.
Also remember that an equilibrium expression is represented by the products divided by the reactants, each of which is raised to its stoichiometric coefficient.
In the case of the reaction given, we can see that it is already balanced. This means we are ready to set up the expression.
Here, the term is being squared because it has a stoichiometric coefficient of
in the equilibrium reaction. Also notice that
does not show up in the expression because it is a solid.
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What is ? Is reaction 3 spontaneous?
There are two concepts to consider in this problem. First, the question asks for the . Reaction 3 has
being reduced so the potential for the half-reaction becomes negative.
half-reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.
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If delta G for a certain reaction is found to be -4955 J/mol, what is the equilibrium constant (Keq) for this reaction?
Given the equation delta G = -RTln(Keq), plugging in the correct values for the variables will lead us to the correct answer. Using e = 2.7, R = 8.314 J/molK, T= 298 K, and G = -4955 J, we can calculate that K = 7.4.
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