Question
\(HBrO(aq)+H2O(l)\)⇄\(H_3O^+(aq)+BrO^−(aq)\) \(K_{eq}=2.8\times 10^{−9}\)
The equilibrium reaction in 0.100M \(HBrO(aq)\) 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?
A The equilibrium concentration of \(BrO^−\) will be much smaller than the equilibrium concentration of \(H_3O^+\) , because \(H_2O\) is the solvent and is present in the largest amount.
B The equilibrium concentration of \(BrO^−\) will be much smaller than the equilibrium concentration of \(HBrO\), because \(K_{eq}\)<<1.
C The equilibrium concentration of \(H_3O^+\) will be much smaller than the equilibrium concentration of \(BrO^−\), because all the \(HBrO\) will react to produce \(BrO^−\).
D The equilibrium concentration of \(H_3O^+\) will be much larger than the equilibrium concentration of \(HBrO\), because \(K_{eq}\)<<1 .
▶️Answer/Explanation
Ans:B
Because \(K_{eq}\)<<1 and \(K_{eq}=\frac{[products]_{eq}}{[reactants]_{eq}}\) , \([BrO^−]_{eq}\)<<\([HBrO]_{eq}\).
Question
\(2SO_2(g)+O_2(g)\)⇄\(2SO_3(g)\) \(K_p\)≈\(2\times 10^5\)
At a certain temperature, \(SO_2(g)\) and \(O_2(g)\) react to produce \(SO_3(g)\) according to the chemical equation shown above. An evacuated rigid vessel is originally filled with \(SO_2(g)\) and \(O_2(g)\), each with a partial pressure of 1atm. Which of the following is closest to the partial pressure of \(O_2(g)\) after the system has reached equilibrium, and why?
A 0atm ; because \(K_p\) is very large, nearly all the \(SO_2(g)\) and \(O_2(g)\) are consumed before the system reaches equilibrium.
B 0.5atm ; because \(K_p\) is very large, nearly all the \(SO_2(g)\) is consumed before the system reaches equilibrium, but an excess amount of \(O_2(g)\) remains at equilibrium.
C 1atm ; because \(K_p\) is very large, the system is already near equilibrium, and there will be very little change to the partial pressure of \(O_2(g)\) .
D 1.5atm ; because \(K_p\) is very large, the decomposition of any \(SO_3(g)\) that forms increases the amount of \(O_2(g)\) at equilibrium.
▶️Answer/Explanation
Ans:B
Based on the large \(K_p\) , the reaction will proceed almost to completion, but based on the stoichiometry, \(SO_2(g)\) is a limiting reactant; thus, the excess \(O_2(g)\) will have partial pressure approximately equal to 0.5atm.
Question
\(Zn(s)+Cu^{2+}(aq)\)⇄\(Zn^{2+}(aq)+Cu(s)\) \(K_{eq}=2\times 10^{37}\)
A 0.10mol sample of solid zinc is added to 500.0mL of 1.0M \(Cu(NO_3)_2(aq)\) . After the mixture sits overnight, which of the following best describes what will most likely be observed and measured the next morning and why?
A Almost all of the \(Zn(s)\) will still be in the beaker with no visible \(Cu(s)\), because the initial mixture is already at equilibrium due to the very large \(K_{eq}\) .
B About half of the \(Zn(s)\) will have disappeared and \(Cu(s)\) will have appeared in the beaker, because the system reaches equilibrium.
C About two-thirds of the \(Zn(s)\) will have disappeared and \(Cu(s)\) will have appeared in the beaker, because the system reaches equilibrium.
D Virtually all of the \(Zn(s)\) will have disappeared and \(Cu(s)\) will have appeared in the beaker, because the reaction proceeds almost to completion at equilibrium due to the very large \(K_{eq}\) .
▶️Answer/Explanation
Ans:D
The reaction should proceed almost to completion at equilibrium because \(K_{eq}\) is very large, resulting in a yield of about 0.10mol of \(Cu(s)\) and a negligible amount of \(Zn(s)\) that would not be visibly detectable.