Exam 13: Chemical Equilibrium

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Consider the equilibrium system Consider the equilibrium system   If C is removed, the equilibrium will ____, and if CO is added, the equilibrium will ____. If C is removed, the equilibrium will ____, and if CO is added, the equilibrium will ____.

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In a balanced chemical equation, if there is no net change in the total number of moles of gas from the reactant side to the product side, then

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If the reaction quotient, Q, is greater than the equilibrium constant, K, then

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Which of the following is not true for equilibrium constants?

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For an exothermic reaction, an increase in temperature

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For the reaction For the reaction  K<sub>c</sub> = 1.37. If the concentrations of A and B are equal, what is the value of that concentration?Kc = 1.37. If the concentrations of A and B are equal, what is the value of that concentration?

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If the value of Kc for a given reaction at a given temperature is a large number, the energy of the products is likely to be ____ the energy of the reactants because ____.

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The equilibrium for a particular chemical reaction is dependent on the

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Consider the reaction  Consider the reaction    for which K<sub>p</sub> = 2.00 at 20<sup> \circ </sup>C. A mixture of NO(g) at a partial pressure of 1.50 atm and NO<sub>2</sub>(g) at a partial pressure of 0.50 atm is allowed to come to equilibrium in a sealed container maintained at 20<sup> \circ </sup>C. What is the total pressure now? The expression for    . for which Kp = 2.00 at 20 ∘\circ C. A mixture of NO(g) at a partial pressure of 1.50 atm and NO2(g) at a partial pressure of 0.50 atm is allowed to come to equilibrium in a sealed container maintained at 20 ∘\circ C. What is the total pressure now? The expression for  Consider the reaction    for which K<sub>p</sub> = 2.00 at 20<sup> \circ </sup>C. A mixture of NO(g) at a partial pressure of 1.50 atm and NO<sub>2</sub>(g) at a partial pressure of 0.50 atm is allowed to come to equilibrium in a sealed container maintained at 20<sup> \circ </sup>C. What is the total pressure now? The expression for    . .

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Consider the equilibrium reaction Consider the equilibrium reaction   A sample of pure NO<sub>2</sub>(g) of concentration 0.140 M is allowed to come to equilibrium. It is then found that 57.0 % of the NO<sub>2</sub>(g) has reacted to form N<sub>2</sub>O<sub>4</sub>(g). What is the value of K<sub>c</sub>? A sample of pure NO2(g) of concentration 0.140 M is allowed to come to equilibrium. It is then found that 57.0 % of the NO2(g) has reacted to form N2O4(g). What is the value of Kc?

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Consider the gas-phase equilibrium A Consider the gas-phase equilibrium A   B. In a series of experiments, different initial amounts of A and B are mixed together, and the mixture in each case allowed to come to equilibrium. Which one of these experiments would yield values for the amounts of A and B present at equilibrium different from all the other experiments? B. In a series of experiments, different initial amounts of A and B are mixed together, and the mixture in each case allowed to come to equilibrium. Which one of these experiments would yield values for the amounts of A and B present at equilibrium different from all the other experiments?

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Which of the following is false?

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Which of the following is true for an equilibrium system?

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A particular reaction mixture (Kp = 10) consisting of both gaseous reactants and products is analyzed and is determined to have a value of Q = 15. The pressure of the reaction mixture is then doubled by the addition of argon gas and the reaction is allowed to proceed to equilibrium. At equilibrium

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Consider the gas-phase equilibrium A Consider the gas-phase equilibrium A   B. Certain amounts of A and B are mixed in a vessel. As they come to equilibrium B. Certain amounts of A and B are mixed in a vessel. As they come to equilibrium

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If a catalyst is added to a chemical reaction, the equilibrium yield of a product will be ____, and the time taken to come to equilibrium will be ____ than before.

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The equilibrium constant expression for the reverse of the reaction given below is The equilibrium constant expression for the reverse of the reaction given below is

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