WorksheetsLe Chatetier's Principle and equilibrium
Total questions: 50
Worksheet time: 1hrs 16mins
What two factors indicate that a reaction is at equilibrium?
Concentrations are equal; the forward and reverse reactions are equal
The concentrations are constant; the forward and reverse reactions are equal
The concentrations are equal; the forward reaction is faster than the reverse reaction
The concentrations are constant; the forward reaction is faster than the reverse reaction
A double sided arrow (↔), indicates that a reaction is: (a)
When the forward and reverse reactions are occurring at the same rate, the reaction is said to be at:
Chemical equilibrium
Chemical reaction
Chemical constant
Chemical peace
If a reversible reaction has reached equilibrium, which of the following is true about the concentration of the reactants vs the products?
Some reactants are present, as are some prooducts
Only reactant is present, there are no products
Only product is present, there are no reactants
Which of the following is true if the value of the equilibrium constant, K, is greater than 1
Products are favored over reactants
Reactants are favored over products
Reactants and products are equally favored
Which of the following is true if the value of the equilibrium constant, K, is less than 1
Products are favored over reactants
Reactants are favored over products
Reactants and products are equally favored
How does a catalyst affect equilibrium?
It makes the reaction occur faster, and makes the equilibrium shift to make more products
It makes the reaction occur faster, but does not shift the placement of the equilibrium
It does not change the rate of the reaction, but makes the equilibrium shift to make more products
A catalyst has no effect on the rate of a reaction or the placement of the equilibrium
Which of the following factors has no effect on the position of the chemical equilibrium for a reaction?
Pressure
Temperature
Catalyst
Concentration
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Adding H2 will cause the equilibrium to:
Shift right
Shift left
Have no change
Speed up
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Increasing the temperature will cause the equilibrium to:
Shift right
Shift left
Have no change
Speed up
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Removing N2 will cause the equilibrium to:
Shift right
Shift left
Have no change
Speed up
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Increasing the volume of the container (decreasing the pressure) will cause the equilibrium to:
Shift right
Shift left
Have no change
Speed up
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Increasing the pressure will cause the equilibrium to:
Shift right
Shift left
Have no change
Speed up
N2(g) + 3H2(g) ↔ 2 NH3(g) + heat
Increasing the pressure causes the equilibrium to shift right. Why?
To increase the pressure more, because there are fewer moles of gas in the products
Because this makes the equilibrium constant, K, increase
To reduce the pressure, because there are fewer moles of gas in the products
To reduce the pressure, because there are more moles of gas in the products
SO2(g) + O2(g) <−> SO3(g)
If the concentration of SO2(g) is increased, the equilibrium of the reaction will ___________.
SO2(g) + O2(g) <−> SO3(g)
If the equilibrium shifts to the right, the concentration of O2(g) will ___________.
energy + N2(g) + O2(g) <−> 2NO(g)
If O2(g) is removed, the concentration of N2 will _______.
Rewrite the above equation with energy as a reactant or product:
heat + N2(g) + O2(g) <−> 2NO(g)
If the heat is removed to the chemical system, the equilibrium will _______.
N2 (g) + 3 H2 (g) <−> 2 NH3 (g)
If the pressure in the system is increased, the reaction will __________________.
ΔHrxn= -453 kJ/mol
If the [B] is decreased then the reaction is will shift to the _______.
2 NO(g) + O2(g) ⇌ 2 NO2(g)
Raising the pressure by lowering the volume of the container will...
Removing O2(g) will
Increasing the pressure on an equilibrium system will
shift the reaction to the side with more moles of chemicals
shift the reaction to the side with more moles of gas
shift the reaction to the side with fewer moles of gas
shift the reaction to the side with fewer moles of chemicals
If an equilibrium constant, K, is 1.8 x 10-5, at equilibrium
I will have more reactants
I will have more products
I will have roughly equal amounts of reactants and products
it is impossible to tell how much reactants and products I will have
For the reaction below, (metane gas reacting with dihydrogen sulfide) which change would cause the equilibrium to shift to the right?
CH4(g) + 2H2S(g) + heat↔ CS2(g) + 4H2(g)
Decrease the concentration of dihydrogen sulfide.
Increase the pressure on the system.
Increase the temperature of the system.
Increase the concentration of carbon disulfide.
Decrease the concentration of methane.
Consider the following reaction:
Fe +3 + SCN ↔ FeSCN 2+
(Light Yellow) (Deep Red)
Adding Fe(NO3)3 produced the following change in the equilibrium:
The color in the test tube became a deeper red color because the equilibrium shifted to make more reactants.
The color in the test tube became a deeper red color because the equilibrium shifted to make more products.
The color in the test tube became a lighter color because the equilibrium shifted to make more reactants.
The color in the test tube became a lighter color because the equilibrium shifted to make more products.
Consider the following reaction:
Fe +3 + SCN ↔ FeSCN 2+
(Light Yellow) (Deep Red)
The reaction you studied became a deeper red color when placed in an ice bath. This means that the reaction is
exothermic.
endothermic.
There is not enough information given to answer the qusestion.
None of the above are correct.
In the reaction, CO (g) + NO2 (g) ↔ CO2 (g) + NO (g), which of the following changes would result in the formation of more products at equilibrium?
increasing the pressure
removing CO (g) from the reaction
adding NO2 (g) to the reaction
adding CO2 to the reaction.
None of the options would result in more products.
SO2 (g) + NO2 (g) <=> SO3 (g) + NO (g)
had reached a state of equilibrium, was found to contain
0.40 mol L-1 SO3 , and 0.30 mol L-1 NO,
0.15 mol L-1NO2 , and 0.20 mol L-1 SO2.
Calculate the equilibrium constant for this reaction.
Fe3O4(s) + 4H2(g) <=> 3Fe(s) + 4H2O(g)
Kc =
What is the concentration equilibrium constant expression?
CO2(g) + H2(g) ↔ CO(g) + H2O(l)
4Br2(g) + CH4(g) ↔ 4HBr(g) + CBr4(g)
The forward reaction points towards the __________.
What is the reaction/equation for this Kc expression
2N2 + O2 ⇌ 2NO
N2 + O2 ⇌ NO
NO ⇌ 2NO
N2 + O2 ⇌ 2NO
What is the equation/reaction for this Kc expression?
O3 ⇌ 3O2
2O3 ⇌ 3O2
2O3 ⇌ O2
3O2 ⇌ 2O3
If Kc is large that means the concentration of the __________ is high and therefore the concencentration of the __________ is low.
reactants, products
species, products
products, reactants
species, reactants
Nothing, without considering the coefficients of the reaction
