progress check 7.6 -->7.8 (10 -->21 )

HBrO(aq) + H₂O(l) ⇄ H₃O⁺(aq) + BrO⁻(aq)   Keq=2.8×10⁻⁹

The equilibrium reaction in 0.100M HBrO(aq) at equilibrium is represented by the equation above. Based on the magnitude of the equilibrium constant, which of the following correctly compares the equilibrium concentrations of substances involved in the reaction, and why?

• The equilibrium concentration of BrO⁻ will be much smaller than the equilibrium concentration of H₃O⁺, because H₂O is the solvent and is present in the largest amount.

• The equilibrium concentration of BrO⁻ will be much smaller than the equilibrium concentration of HBrO, because Keq<<1

• The equilibrium concentration of H₃O⁺ will be much smaller than the equilibrium concentration of BrO⁻, because all the HBrO will react to produce BrO⁻

• The equilibrium concentration of H₃O⁺ will be much larger than the equilibrium concentration of HBrO, because Keq<<1.

At a certain temperature, SO₂(g) and O₂(g) react to produce SO₃(g) according to the chemical equation shown above. An evacuated rigid vessel is originally filled with SO₂(g) and O₂(g), each with a partial pressure of 1atm

1atm. Which of the following is closest to the partial pressure of O₂(g) after the system has reached equilibrium, and why?

• 0atm; because Kp is very large, nearly all the SO₂(g) and O₂(g) are consumed before the system reaches equilibrium.

• 0.5atm; because Kp is very large, nearly all the SO₂(g) is consumed before the system reaches equilibrium, but an excess amount of O₂(g) remains at equilibrium.

• 1atm; because Kp is very large, the system is already near equilibrium, and there will be very little change to the partial pressure of O₂(g).

• 1.5atm; because KpKp is very large, the decomposition of any SO3(g) that forms increases the amount of O₂(g) at equilibrium.

A 0.10mol sample of solid zinc was added to 500.0mL

500.0mL of 1.0M Cu(NO3)2(aq). After the mixture sits overnight, which of the following best describes what was most likely observed and measured the next morning and why?

• Almost all of the Zn(s) will still be in the beaker with no visible Cu(s), because equilibrium was not reached due to the very large Keq.

• About half of the Zn(s) will have disappeared and Cu(s) will have appeared in the beaker, because the system reached equilibrium.

• About two-thirds of the Zn(s) had disappeared and Cu(s) will have appeared in the beaker, because the system reached equilibrium.

• Virtually all of the Zn(s) will have disappeared and Cu(s) will have appeared in the beaker, because the reaction went almost to completion at equilibrium due to the very large Keq.

• $K_3=K_1\times\frac{1}{K_2}$

• $K_3=\frac{1}{K_1}\times K_2$

• $K_3=K_1-K_2$

• $K_3=-K_1+K_2$

• $K_{eq}=K_1+\left(2\times K_2\right)$

• $K_{eq}=K_1-\left(2\times K_2\right)$

• $K_{eq}=K_1\times\left(K_2\right)^2$

• $K_{eq}=\frac{K_1}{\left(K_2\right)^2}$

In a large reaction vessel at a constant temperature, nitrogen dioxide and dinitrogen tetroxide are in a state of dynamic equilibrium, as represented by the chemical equations shown above. The equilibrium constants for the reactions are K₁ and K₂. Which of the following quantities can most easily be used to find the value of K₂?

• The value of ΔH for the reaction

• The temperature of the system

• The volume of the system

• The value of K₁

• The color of the solution changed from blue to pink, because lowering the temperature increased the collision frequency between CoCl₄²⁻ and H₂O.

• The color of the solution changed from blue to pink, because cooling caused the equilibrium to shift to form the pink-colored Co(H₂O)₆²⁺.

• The color of the solution did not change, because more water was not added to the solution.

• The color of the solution did not change, because heat is not released from this reaction.

• The color of the solution will lighten because most of the volume added is solvent.

• The color of the solution will darken because the equilibrium will favor the dissociation of FeSCN²⁺.

• The color of the solution will darken because the equilibrium will favor the formation of more FeSCN²⁺.

• The color of the solution will not change because the solution already contains Fe³⁺ ions

A chemist carried out the reaction at 573K, starting with 0.100mol of each reactant in a 1.00L container with variable volume. The reaction mixture quickly reached equilibrium, as indicated in the graph above. The chemist reduces the volume of the equilibrium system by half while keeping the temperature constant. Which of the following predictions about the yield of the reaction is best, and why?

• By halving the volume, the pressure doubles. The system will respond to the increase in pressure by decreasing the total number of moles of gas in the system. Thus, the yield will increase because the reaction will shift toward more product.

• By halving the volume, the pressure doubles. The system will respond to the increase in pressure by increasing the total number of moles of gas in the system. Thus, the yield will decrease because the reaction will shift toward more reactants.

• By halving the volume, the pressure is reduced by half. The system will respond to the decrease in pressure by increasing the total number of moles of gas in the system. Thus, the yield will decrease because the reaction will shift toward more reactants.

• By halving the volume, the pressure is reduced by half. The system will respond to the decrease in pressure by decreasing the total number of moles of gas in the system. Thus, the yield will increase because the reaction will shift toward more products.

• The rate of the reverse reaction will be less than the rate of the forward reaction and additional CH₃OH(g) will be consumed, because Kc<Qc.

• The rate of the reverse reaction will be greater than the rate of the forward reaction and additional CH₃OH(g) will be consumed because Kc<Qc.

• The rate of the forward reaction will be less than the rate of the reverse reaction and additional CH₃OH(g) will be produced, because Kc<Qc.

• The rate of the forward reaction will be greater than the rate of the reverse reaction and additional CH₃OH(g) will be produced, because Kc<Qc.