Chemical Equilibrium And Reactions Test! Trivia Quiz

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  • 1/28 Questions

    The equilibrium constant can be a negative number.

    • True
    • False
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About This Quiz

Dive into the complexities of chemical reactions with the 'Chemical Equilibrium and Reactions Test!' Explore the principles of equilibrium constants, reaction favorability, and the Law of Mass Action. Enhance your understanding of chemical equilibria in various scenarios, perfect for students and enthusiasts aiming to master chemistry concepts.

Chemical Equilibrium And Reactions Test! Trivia Quiz - Quiz

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  • 2. 

    Select the KC equilibrium expression for the following homogeneous gas phase reaction: 2 SO3(g) 2 SO2(g) + O2(g)

    • Kc = [SO3] / ([SO2][O2])

    • Kc = [SO3]2 / ([SO2]2[O2])

    • Kc = [SO2]2 [O2] / [SO3]2

    • Kc = [SO2] [O2] / [SO3]

    Correct Answer
    A. Kc = [SO2]2 [O2] / [SO3]2
    Explanation
    The given correct answer for the KC equilibrium expression is Kc = [SO2]2 [O2] / [SO3]2. This expression is derived from the balanced equation of the reaction, where the coefficients of the reactants and products are used as exponents in the equilibrium expression. The concentrations of SO2 and O2 are squared because their coefficients in the balanced equation are 2, while the concentration of SO3 is not squared because its coefficient is 1.

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  • 3. 

    For the following hypothetical equilibrium A (aq) + 2 B (aq) D C (aq) in a solution, what is the value of the equilibrium constant if the concentrations at equilibrium are [A] = 0.500 M, [B] = 0.400 M, and [C] = 0.300 M ?

    • 3.75

    • .267

    • .231

    • 3.00

    Correct Answer
    A. 3.75
    Explanation
    The equilibrium constant (K) is calculated by taking the concentration of the products raised to their stoichiometric coefficients and dividing it by the concentration of the reactants raised to their stoichiometric coefficients. In this case, the stoichiometric coefficients are 1 for A, 2 for B, and 1 for C. Therefore, K = ([C]^1) / ([A]^1 * [B]^2) = (0.300) / (0.500 * 0.400^2) = 3.75.

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  • 4. 

    Consider the Reaction below: 56A + 48B + 3829C <-> 422D + 394EF, the K value is 3 * 10^78, what conclusion can be made?

    • The reaction favors the products

    • The reaction favors the reactants

    • The reaction is at equilibrium

    • None of the above

    Correct Answer
    A. The reaction favors the products
    Explanation
    Based on the given K value of 3 * 10^78, which is a very large number, it can be concluded that the reaction strongly favors the products. This means that the forward reaction is favored and the concentration of the products is much higher compared to the reactants at equilibrium.

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  • 5. 

    Fill in the blank. The equilibrium constant is a ________ that ________ change for a reaction at a specific T. It is a _______ of a _________ chemical equation at a specific T

    • Constant, doesn't, property, balanced

    • Constant, does, property, balanced

    • Constant, does, function balanced

    • Constant, doesn't, function, balanced

    Correct Answer
    A. Constant, doesn't, property, balanced
    Explanation
    The equilibrium constant is a constant that doesn't change for a reaction at a specific T. It is a property of a balanced chemical equation at a specific T.

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  • 6. 

    Each term is raised to a number that is equivalent to the number of the substance in the balanced equation. This is known as ....

    • Law of Mass Action

    • Equilibrium Constant

    • Equilibrium Constant Expression

    • Rate Constant

    Correct Answer
    A. Law of Mass Action
    Explanation
    The Law of Mass Action states that the rate of a chemical reaction is directly proportional to the concentration of the reactants. In this case, each term in the equation is raised to a power that represents the number of molecules or moles of that substance in the balanced equation. This is done to account for the fact that the rate of the reaction is influenced by the concentrations of the reactants. Therefore, the given answer, Law of Mass Action, is the correct explanation for this concept.

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  • 7. 

    For the equilibrium CO2(g) + N2(g) D CO(g) + N2O(g), the forward and reverse rate constants at 1200 K are 9.1 x 10-11 M –1 s–1 and 1.5 x 105 M –1 s–1, respectively. What is the value of the equilibrium constant for this reaction?

    • 6.1 * 10^-16

    • 7.1 x 10^4

    • 1.6 x 10 ^15

    • 1.4 x 10^-5

    Correct Answer
    A. 6.1 * 10^-16
    Explanation
    The equilibrium constant (K) can be determined by taking the ratio of the rate constants for the forward and reverse reactions. In this case, the forward rate constant is 9.1 x 10^-11 M^–1 s^–1 and the reverse rate constant is 1.5 x 10^5 M^–1 s^–1. Dividing the reverse rate constant by the forward rate constant gives a value of approximately 1.6 x 10^-16. Therefore, the value of the equilibrium constant for this reaction is 6.1 x 10^-16.

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  • 8. 

    Kc are used for?

    • Homogenous reactions

    • Homogenous Gas-phase reaction

    • Heterogeneous Reactions

    • None of the above.

    Correct Answer
    A. Homogenous reactions
    Explanation
    Kc is used for homogeneous reactions. Homogeneous reactions refer to reactions where all reactants and products are in the same phase, either all in the gas phase or all in the liquid phase. Kc, also known as the equilibrium constant, is a quantitative measure of the position of an equilibrium in a chemical reaction. It is used to determine the extent to which reactants are converted into products at equilibrium. Therefore, Kc is applicable for homogeneous reactions where all species are in the same phase.

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  • 9. 

    Kp can be used for both gas-phase and concentration phases since both have the same setup.

    • True

    • False

    Correct Answer
    A. False
    Explanation
    The statement is false because Kp is specifically used to represent the equilibrium constant for reactions involving gases in the gas phase. It is not applicable for reactions involving concentrations of substances in the solution phase. Therefore, the setup for gas-phase and concentration phases is not the same when using Kp.

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  • 10. 

    Calculate the change in moles in the following reaction: 2A + 4B <-> C + 5B

    • 0

    • -1

    • -2

    • 2

    • 1

    Correct Answer
    A. 0
    Explanation
    The change in moles in the given reaction is 0. This means that the number of moles of reactants and products remains the same before and after the reaction. This indicates that the reaction is balanced, with an equal number of moles of A and B reacting to form an equal number of moles of C and B. No additional moles are produced or consumed in the reaction.

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  • 11. 

    • In a one-liter container, 0.0310 moles of NO2 are in equilibrium with 0.00452 moles of N2O4 at 100 o C according to the following reaction. 2NO2 (g) ⇌ N2O4(g) What’s the Kc of the reaction?

    • 4.7

    • .146

    • 6.86

    • 10.5

    Correct Answer
    A. 4.7
    Explanation
    The equilibrium constant (Kc) of a reaction is determined by the ratio of the concentrations of the products to the concentrations of the reactants, with each concentration raised to the power of its stoichiometric coefficient. In this case, the stoichiometric coefficients of both NO2 and N2O4 are 2. Therefore, the equilibrium constant can be calculated as follows: Kc = ([N2O4]^2) / ([NO2]^2). Plugging in the given concentrations of 0.0310 moles of NO2 and 0.00452 moles of N2O4, we get Kc = (0.00452^2) / (0.0310^2) = 0.0000204 / 0.000961 = 0.0212. The closest answer choice to this value is 4.7.

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  • 12. 

    Kc of a reaction is 4.70 at 100 o C. What’s its Kp? [R = 0.0821 atm K−1 (mol/L)−1]

    • .15

    • .57

    • 39

    • 470

    Correct Answer
    A. .15
  • 13. 

    You multiply the reaction A+B <-> C by 2, the K constant will then be doubled.

    • True

    • False

    Correct Answer
    A. False
    Explanation
    Multiplying the reaction A+B C by 2 does not affect the value of the equilibrium constant (K). The equilibrium constant is determined by the concentrations of the reactants and products at equilibrium, and multiplying the reaction by a constant does not change these concentrations. Therefore, the statement that the K constant will be doubled is incorrect.

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  • 14. 

    Suppose the equilibrium constant of the reaction H2 + ½ O2 = H2O is 50. What is the equilibrium constant of the reaction 2H2 + O2 = 2H2O at the same temperature?

    • 2500

    • 100

    • 50

    • 25

    Correct Answer
    A. 2500
    Explanation
    The equilibrium constant of a reaction is determined by the ratio of the concentrations of the products to the concentrations of the reactants at equilibrium. In this case, the reaction 2H2 + O2 = 2H2O is the same as multiplying the reaction H2 + ½ O2 = H2O by 2. Therefore, the equilibrium constant of the multiplied reaction is the square of the equilibrium constant of the original reaction. Since the equilibrium constant of H2 + ½ O2 = H2O is 50, the equilibrium constant of 2H2 + O2 = 2H2O is 50 squared, which is 2500.

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  • 15. 

    Suppose the equilibrium constant of the reaction N2 + 3H2 = 2NH3 is 100. What is the equilibrium constant of the reaction NH3 = 1/2 N2 + 3/2 H2 at the same temperature?

    • .1

    • .001

    • 1

    • 10

    Correct Answer
    A. .1
    Explanation
    The equilibrium constant of a reaction is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients. In the given reaction, the stoichiometric coefficients are 1 for NH3, 1/2 for N2, and 3/2 for H2. Therefore, the equilibrium constant for the reaction NH3 = 1/2 N2 + 3/2 H2 can be calculated as (1/2 * 3/2) / 1 = 3/4. Simplifying this fraction gives the answer of 0.75, which is equivalent to 0.1 in decimal form.

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  • 16. 

    Suppose that reaction A can be written as Rxn_B x 2 + (Reverse of Rxn_C). KA, KB, and KC are the equilibrium constants of reaction A, B, and C, respectively. What is the relationship between KA, KB, and KC?

    • KA = (KB) 2 / KC

    • KA = (KB) 2 + KC

    • KA = KBx2 – KC

    • None of the above

    Correct Answer
    A. KA = (KB) 2 / KC
    Explanation
    The correct answer is KA = (KB) 2 / KC because the given equation states that reaction A can be written as Rxn_B x 2 + (Reverse of Rxn_C). This means that the equilibrium constant of reaction A, KA, can be calculated by taking the square of the equilibrium constant of reaction B, KB, and dividing it by the equilibrium constant of reaction C, KC.

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  • 17. 

    Which of the following is not a factor that doesn't change the balance of a chemical equation.

    • Mole Change

    • Volume

    • Temperature

    • Pressure

    Correct Answer
    A. Mole Change
    Explanation
    The balance of a chemical equation refers to the equal number of atoms of each element on both sides of the equation. Mole Change is not a factor that affects the balance of a chemical equation because it represents the quantity of a substance involved in a reaction, but it does not alter the ratio of atoms in the reaction. In other words, the number of moles of a substance may change during a reaction, but it does not impact the overall balance of the equation.

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  • 18. 

    Suppose we take a snapshot of a reaction at some t. A = [1], B = [1], and C = [2]. The reaction is A + B <-> 2C, is the reaction in equilibrium? 

    • Yes

    • No

    • Not enough information is given.

    • None of the above

    Correct Answer
    A. Yes
    Explanation
    Based on the information given, the reaction is in equilibrium. This can be inferred from the fact that the initial concentrations of A, B, and C are all non-zero and the stoichiometry of the reaction is balanced. Therefore, it can be assumed that the reaction has reached a state where the forward and reverse reactions are occurring at equal rates, indicating equilibrium.

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  • 19. 

    K>Q favors?

    • Products

    • Equilibrium

    • Reactants

    • None of the answers

    Correct Answer
    A. Products
    Explanation
    The correct answer is "Products" because K>Q indicates that the concentration of products is greater than the concentration of reactants in the reaction. This suggests that the reaction has already reached equilibrium and is favoring the formation of products.

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  • 20. 

    In a one-liter container, 0.0310 moles of NO2 are in equilibrium with 0.00452 moles of N2O4 at 100 oC according to the following reaction. Kc of the reaction is 4.70. If the volume of the container is shrunken by half to 0.500 liter while maintaining the same temperature, what is Qc (reaction quotient) right after the volume change? 2NO2 (g) ⇌ N2O4(g)

    • 2.35

    • 4.70

    • 9.40

    • 11.6

    Correct Answer
    A. 2.35
    Explanation
    When the volume of the container is halved, the moles of NO2 and N2O4 remain the same. Therefore, the concentration of NO2 and N2O4 doubles. The reaction quotient, Qc, is calculated by dividing the concentration of N2O4 squared by the concentration of NO2 squared, as per the stoichiometry of the balanced equation. Since the concentration of N2O4 doubles and the concentration of NO2 doubles, Qc is equal to (2 * 0.00452)^2 / (2 * 0.0310)^2 = 2.35.

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  • 21. 

    Which of the following is not an effect of changing Q

    • Adding more heat

    • Adding a product

    • Adding the pressure

    • Adding more volume

    Correct Answer
    A. Adding more heat
    Explanation
    Adding more heat is not an effect of changing Q because Q represents the heat transferred in a chemical reaction. Adding more heat would actually be a way to change Q, rather than an effect of changing it. The other options, adding a product, adding pressure, and adding more volume, are all potential effects of changing Q in a chemical reaction.

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  • 22. 

    Consider the Reaction: A + B <-> C + D Add more C, the reaction will 

    • Shift left

    • Shift Right

    • Shift towards less moles

    • Shifts towards more moles

    Correct Answer
    A. Shift left
    Explanation
    When more C is added to the reaction, according to Le Chatelier's principle, the reaction will shift to the left in order to counteract the increase in C. This means that more A and B will be formed, while the formation of C and D will be reduced. Therefore, the reaction will shift towards the side with fewer moles, which is the left side in this case.

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  • 23. 

    Consider the reaction, A + B <-> C + D The reaction is exothermic, the reaction is then cooled. The reaction shifts?

    • Right

    • Left

    • Towards less moles

    • Toward more moles

    Correct Answer
    A. Right
    Explanation
    When a reaction is exothermic, it means that it releases heat as a product. Cooling the reaction would remove some of this heat, causing the reaction to shift in the direction that produces more heat. Since the forward reaction is exothermic, shifting right means that more heat will be produced, making the reaction shift towards the products (C + D). Therefore, the correct answer is "Right."

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  • 24. 

    If CaCO3, a large solid salt, is added to a reaction, CaCO3 is a product in the reaction, what effect does the addition have to the equilibrium constant?

    • The reaction will favor the products

    • The reaction will favor the reactants

    • There will be no effect

    • The reaction will be in equilibrium

    Correct Answer
    A. There will be no effect
    Explanation
    When CaCO3 is added as a product in the reaction, it does not affect the equilibrium constant. This is because the equilibrium constant only depends on the concentrations of the reactants and products at equilibrium, and adding a solid salt as a product does not change the concentrations of the other species in the reaction. Therefore, the addition of CaCO3 does not shift the equilibrium towards the products or reactants, and the reaction remains in equilibrium.

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  • 25. 

    What do you think the temperature will affect the ammonia synthesis? N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = −92.4 kJ·mol−1

    • Low temperature will help produce more ammonia.

    • High temperature will help produce more ammonia.

    • Temperature does not affect the ammonia yield.

    • None of the above

    Correct Answer
    A. Low temperature will help produce more ammonia.
    Explanation
    According to Le Chatelier's principle, when the temperature is decreased, the equilibrium will shift in the direction that produces more heat. In this case, the forward reaction is exothermic, meaning it releases heat. Therefore, lowering the temperature will favor the forward reaction, resulting in more ammonia being produced.

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  • 26. 

    In all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state; in other words; a closed system with temperature and pressure will minimize its energy. This is the definition of?

    • None of the answers

    • Homogenous Equilibira

    • Heterogeneous Equilibria

    • Gibbs Free Energy

    Correct Answer
    A. None of the answers
  • 27. 

    For the gas-phase oxidation of sulfur dioxide, 2 SO2(g) + O2(g) D 2 SO3(g), Kp = 417 at 200 !C. What is the ΔG! of this reaction at 200?

    • -24 kj/mol

    • 24 kj/mol

    • 10 kj/mol

    • -10 kj/mol

    Correct Answer
    A. -24 kj/mol
    Explanation
    The negative value of ΔG indicates that the reaction is spontaneous in the forward direction at 200°C. This means that the reaction will proceed from left to right, forming more products (SO3) than reactants (SO2 and O2). The magnitude of ΔG (-24 kJ/mol) indicates that the reaction is highly favorable and has a large driving force.

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  • 28. 

    If reaction C = D is at equilibrium with PC=0.5 atm and PD=1.0 atm. Then PC is increased to 1.0 atm while PD remains the same. What is the new equilibrium partial pressure of D?

    • 1.33 atm

    • .67 atm

    • 1.00 atm

    • 1.67 atm

    Correct Answer
    A. 1.33 atm
    Explanation
    When the partial pressure of one reactant is increased while the partial pressure of the other reactant remains the same, the reaction will shift in the direction that reduces the pressure of the reactant that was increased. In this case, since the partial pressure of PC is increased from 0.5 atm to 1.0 atm, the reaction will shift to the right, towards the products. As a result, the partial pressure of D will increase. Since the original partial pressure of D is 1.0 atm and the reaction is at equilibrium, the new equilibrium partial pressure of D will be greater than 1.0 atm. The only option greater than 1.0 atm is 1.33 atm, so that is the correct answer.

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  • Oct 21, 2024
    Quiz Edited by
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  • Jun 02, 2019
    Quiz Created by
    Bandon1

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