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Chapter 1: Problem 20

$$\mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \leftrightarrow 2 \mathrm{IBr}(g)$$ At \(150^{\circ} \mathrm{C},\) the equilibrium constant, \(K_{c},\) for the reaction shown above has a value of \(300 .\) This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to \(\operatorname{Br}_{2}(g)\) and \(I_{2}(g)\) in the container? \(\begin{array}{lll}{} & {\operatorname{Br}_{2}(g)} & {\mathrm{I}_{2}(g)} \\\ {\text { (A) }} & {0.1 \mathrm{atm}} & {0.3 \mathrm{atm}} \\ {\text { (B) }} & {0.3 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (C) }} & {1 \mathrm{atm}} & {1 \mathrm{atm}} \\ {\text { (D) }} & {1 \mathrm{atm}} & {3 \mathrm{atm}}\end{array}\)

Short Answer

Expert verified
The correct answer is (A): the partial pressures of \(Br_{2}(g)\) and \(I_{2}(g)\) could be 0.1 atm and 0.3 atm, respectively.

Step by step solution

01

Define the equilibrium expression based on the balanced chemical reaction

The balanced reaction is \(Br_{2}(g) + I_{2}(g) \leftrightarrow 2 IBr(g)\). Thus the equilibrium expression is: \(K_c = \frac{[IBr]^2}{[Br_{2}][I_{2}]}\), where [X] represents the partial pressure of X in the equilibrium mixture.
02

Substitute given values into the equilibrium expression

The problem provides that \(K_c = 300\) and [IBr] = 3 atm. Substitute these values into the equilibrium expression: \(300 = \frac{3^{2}}{[Br_{2}][I_{2}]}\).
03

Solve for [Br_{2}][I_{2}]

Calculating the equation from Step 2 gives: [Br_{2}][I_{2}] = \frac{3^{2}}{300} = 0.03 atm^2.
04

Analyze and match the solution to the provided choices

The product of the partial pressures of \(Br_{2}(g)\) and \(I_{2}(g)\) must equal 0.03 atm^2. The only choice that satisfies this is choice (A): [Br_{2}] = 0.1 atm and [I_{2}] = 0.3 atm, since 0.1 atm*0.3 atm equals 0.03 atm^2.

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Most popular questions from this chapter

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. At the equivalence point, is the solution acidic, basic, or neutral? Why? (A) Acidic; the strong acid dissociates more than the weak base (B) Basic; the only ion present at equilibrium is the conjugate base (C) Basic; the higher concentration of the base is the determining factor (D) Neutral; equal moles of both acid and base are present

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ As the reaction progresses, what will happen to the overall voltage of the cell? (A) It will increase as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (B) It will increase as \(\left[\mathrm{Cu}^{+}\right]\) increases. (C) It will decrease as \(\left[\mathrm{Zn}^{2+}\right]\) increases. (D) The voltage will remain constant.

In a voltaic cell with a Cu(s) \(| \mathrm{Cu}^{2+}\) cathode and \(\mathrm{a} \mathrm{Pb}^{2+} | \mathrm{Pb}\) (s) anode, increasing the concentration of \(\mathrm{Pb}^{2+}\) causes the voltage to decrease. What is the reason for this? (A) The value for Q will increase, causing the cell to come closer to equilibrium. (B) The solution at the anode becomes more positively charged, leading to a reduced electron flow. (C) The reaction will shift to the right, causing a decrease in favor ability. (D) Cell potential will always decrease anytime the concentration of any aqueous species present increases.

A strong acid/strong base titration is completed using an indicator which changes color at the exact equivalence point of the titration. The protonated form of the indicator is HIn, and the deprotonated form is In. At the equivalence point of the reaction: (A) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]\) (B) \([\mathrm{HIn}]=1 /\left[\mathrm{In}^{-}\right]\) (C) \([\mathrm{HIn}]=2\left[\mathrm{In}^{-}\right]\) (D) \([\mathrm{HIn}]=\left[\mathrm{In}^{-}\right]^{2}\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ What will happen in the salt bridge as the reaction progresses? (A) The Na' ions will flow to the Cu/Cu' half-cell. (B) The Br' ions will flow to the Cu/Cu' half-cell. (C) Electrons will transfer from the Cu/Cu' half-cell to the Zn/Zn" half cell. (D) Electrons will transfer from the \(\mathrm{Zn} / \mathrm{Zn}^{2+}\) half- cell to the Cu/Cu' half cell.
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