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Chapter 19: Problem 9

Consider a reaction $\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \rightleftharpoons 2 \mathrm{AB}(g),$ with atoms of A shown in red in the diagram and atoms of \(\mathrm{B}\) shown in blue. (a) If \(K_{\mathrm{c}}=1,\) which box represents the system at equilibrium? (b) If \(K_{\mathrm{c}}=1,\) which box represents the system at \(Q < K_{\mathrm{c}} ?(\mathbf{c})\) Rank the boxes in order of increasing magnitude of \(\Delta G\) for the reaction. [ Sections 19.5 and 19.7\(]\)

Short Answer

Expert verified
The short version of the answer is: a) The box representing the system at equilibrium is the one that satisfies the equation: \(1 = \frac{([AB]_0 + 2x)^2}{([A_{2}]_0 - x)([B_{2}]_0 - x)}\) b) The box representing the system with \(Q < K_{c}\) will be the one with a Q value less than 1. c) The boxes can be ranked in order of increasing magnitude of ΔG by analyzing the relationship between Q and Kc: 1. If Q < Kc, ΔG > 0. 2. If Q = Kc, ΔG = 0. 3. If Q > Kc, ΔG < 0.

Step by step solution

01

Calculate the concentration of each species at equilibrium

Since the graphical representation of the system at equilibrium is not known, we will assume the generic scenario for each box. Let the initial concentrations of A2, B2, and AB be [A2]0, [B2]0, and [AB]0, respectively. At equilibrium, the concentrations would be [A2], [B2], and [AB]. Let x mol of A2 and B2 react to produce 2x mol of AB. Then, the equilibrium concentrations are: [A2] = [A2]0 - x [B2] = [B2]0 - x [AB] = [AB]0 + 2x
02

Write the expression for Kc

Kc is the equilibrium constant in terms of concentration. For the given reaction, we can write the expression for Kc as: \(K_{c} = \frac{[AB]^2}{[A_{2}][B_{2}]}\)
03

Substitute the concentrations in the Kc expression

Now, using the equilibrium concentrations in terms of x, the Kc expression becomes: \[K_{c} = \frac{([AB]_0 + 2x)^2}{([A_{2}]_0 - x)([B_{2}]_0 - x)}\] Given that \(K_{c} = 1\), we have: \[1 = \frac{([AB]_0 + 2x)^2}{([A_{2}]_0 - x)([B_{2}]_0 - x)}\] We need to find out the box that satisfies this equation, which represents the system at equilibrium. #b) Find the box representing the system Q<Kc.#
04

Write the expression for Q

The reaction quotient Q is also expressed as: \(Q = \frac{[AB]^2}{[A_{2}][B_{2}]}\) However, unlike Kc, Q represents the current state of the system, not necessarily the equilibrium state.
05

Compare Q and Kc

We need to find the box where \(Q < K_{c}\). Since we are given that \(K_{c} = 1\), the system with \(Q < K_{c}\), will be the box where Q value is less than 1. #c) Rank the boxes in order of increasing ΔG#
06

Relate ΔG to Q and Kc

The relationship between ΔG, Q, and Kc is given by the equation: \[\Delta G = \Delta G^{\circ} + RT \ln {Q}\] At equilibrium, ΔG = 0, and thus: \[\Delta G^{\circ} = - RT \ln {K_{c}}\] So, the equation can be rewritten as: \[\Delta G = RT (\ln {Q} - \ln {K_{c}}) = RT \ln{\frac{Q}{K_{c}}}\]
07

Rank the boxes based on ΔG

For ranking the boxes according to increasing ΔG values, we need to analyze the relationship between Q and Kc, and how it affects ΔG: 1. If Q < Kc, then \(Q/K_{c} < 1\) and ΔG > 0. 2. If Q = Kc, then \(Q/K_{c} = 1\) and ΔG = 0. 3. If Q > Kc, then \(Q/K_{c} > 1\), and ΔG < 0. We can rank the boxes based on these relationships between Q, Kc, and ΔG.

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Most popular questions from this chapter

Using \(S^{\circ}\) values from Appendix C, calculate \(\Delta S^{\circ}\) values for the following reactions. In each case account for the sign of \(\Delta S^{\circ} .\) (a) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) (c) \(\mathrm{Be}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BeO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\)

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