AP Chemistry: 7.4 Calculating the Equilibrium Constant – Exam Style questions with Answer- MCQ

Question

In an experiment involving the reaction shown above, a sample of pure HI was placed inside a rigid container at a certain temperature. The table above provides the initial and equilibrium concentrations for some of the substances in the reaction. Based on the data, which of the following is the value of the equilibrium constant \(K_{eq}\) for the reaction, and why?

A \(K_{eq}=2.5\times 10^{−1}\) , because \([I_2]_{eq}=2\times [HI]_{eq}\).

B \(K_{eq}=6.3\times 10^{−2}\) , because \([I_2]_{eq}=\frac{1}{2}[HI]_{eq}\).

C \(K_{eq}=1.6\times 10^{−2}\) , because \([I_2]_{eq}=[HI]_{eq}\).

D \(K_{eq}=3.1\times 10^{−2}\) , because \([I_2]_{eq}=2\times [HI]_{eq}\).

▶️Answer/Explanation

Ans:C

The mathematical expression for the equilibrium constant is \(K_{eq}=\frac{[H_2]_{eq}[I_2]_{eq}}{[HI][HI]^{2}_{eq}}\) . Since \(H_2\) and \(I_2\) are both products and have the same stoichiometric coefficient, their equilibrium concentrations will be the same. Therefore, \(K_{eq}=\frac{(0.030)(0.030)}{(0.24)^2}=1.6\times 10^{−2}\).

Question

                                                   

Which of the following indicates the equilibrium value of \(PCl_2\) and the approximate value of the equilibrium constant, \(K_p\)?

A \(PCl_2=0.10\) atm and \(K_p=5.0\).

B \(PCl_2=0.20\) atm and \(K_p=0.050\).

C \(PCl_2=0.20\) atm and \(K_p=20\).

D \(PCl_2=0.80\) atm and \(K_p=0.025\).

▶️Answer/Explanation

Ans: C

In the balanced equation for the reaction, the coefficients of \(SO_2\) and \(Cl_2\) are both 1. Therefore, the number of atmospheres of \(Cl_2\) that reacted must be the same as the number of atmospheres of \(SO_2\) that reacted. Since the equilibrium partial pressure of \(SO_2\) is 0.20atm, the equilibrium partial pressure of \(Cl_2\) must be 0.20atm. For the reaction,

\(K_p=\frac{(PSO_2Cl_2)(PSO_2)}{(PCl_2)}\); thus, the value of \(K_p=\frac{0.80}{(0.20)(0.20)}=20\).

Question

                                   

The diagram above represents the equilibrium between two isomers of 2-butene. The equilibrium constant, \(K_c\) , is 1.2 at a certain temperature. Two identical vessels each contain an equilibrium mixture of the two gases at that temperature. The concentration of cis‑2‑butene in the second vessel is twice the concentration in the first vessel. What is the concentration of trans‑2‑butene in the second vessel compared to that in the first vessel?

A Half the concentration of that in the first vessel

B The same concentration as that in the first vessel

C Twice the concentration of that in the first vessel

D Four times the concentration of that in the first vessel

▶️Answer/Explanation

Ans:C

\(K_c=\frac{[trans-2-butene]}{[cis-2-butene]}\). \(K_c\) is constant at constant temperature, so if \([cis-2-butene]\) doubles, \([trans-2-butene]\) must double.

Question

Based on the information above, which of the following expressions represents the equilibrium constant, K, for the reaction represented by the equation below?

\(La^{3+}+CO_3^{2-}\rightleftharpoons LaCO_3^+\)

(A)\(K=(K_1)(K_a)(K_w)\)

(B)\(K=\frac{(K_1)(K_a)}{(K_w)}\)

(C)\(K=\frac{(K_1)}{(K_a)(K_w)}\)

(D)\(K=\frac{(K_1)(K_w)}{(K_a)}\)

▶️Answer/Explanation

Ans:D

Question

The value of the equilibrium constant for the reaction represented above is \(1 × 10^{10}\). What is the value of the equilibrium constant for the following reaction?

\(2Ge(g)Cl_{4}(g)\rightleftharpoons 2Ge(g)+4Cl_{2}(g)\)

(A) 1 × \(10^{-20}\)
(B) 1 × \(10^{-10}\)
(C) 1 × \(10^{10}\)
(D) 1 ×\( 10^{20}\)

▶️Answer/Explanation

Ans:A

Scroll to Top