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Chapter 12: Problem 31

A compound, \(\mathrm{X}\), decomposes at \(131^{\circ} \mathrm{C}\) according to the following equation: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ If a flask initially contains \(\mathrm{X}, \mathrm{A}\), and \(\mathrm{C}\), all at partial pressures of \(0.250 \mathrm{~atm}\), in which direction will the reaction proceed?

Short Answer

Expert verified
Answer: The reaction will proceed in the reverse direction to reach equilibrium.

Step by step solution

01

Understand the reaction and write the equilibrium constant expression.

Given the reaction: $$2 \mathrm{X}(g) \rightleftharpoons \mathrm{A}(g)+3 \mathrm{C}(g) \quad K=1.1 \times 10^{-3}$$ We can write the equilibrium constant expression as: $$K = \frac{[\mathrm{A}][\mathrm{C}]^3}{[\mathrm{X}]^2}$$
02

Calculate the initial reaction quotient (Q).

Using the initial partial pressures of X, A, and C, we can calculate the initial reaction quotient: $$Q = \frac{[\mathrm{A}][\mathrm{C}]^3}{[\mathrm{X}]^2} = \frac{(0.250)(0.250)^3}{(0.250)^2}$$
03

Simplify the expression for the reaction quotient (Q).

After simplification, we have: $$Q=\frac{(0.250)^4}{(0.250)^2} = (0.250)^2$$
04

Evaluate the reaction quotient (Q) and compare to the equilibrium constant (K).

We have: $$Q = (0.250)^2 = 0.0625$$ Now compare Q to K: $$Q > K \Rightarrow (0.0625 > 1.1 \times 10^{-3})$$
05

Determine the direction of the reaction.

Since Q is greater than K, the reaction will proceed in the reverse direction to reach the equilibrium. The decomposition of X will decrease, and the formation of A and C will increase until Q equals to K, reaching the equilibrium.

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Most popular questions from this chapter

Given the following descriptions of reversible reactions, write a balanced equation (simplest whole-number coefficients) and the equilibrium constant expression \((K)\) for each. (a) Nitrogen gas reacts with solid sodium carbonate and solid carbon to produce carbon monoxide gas and solid sodium cyanide. (b) Solid magnesium nitride reacts with water vapor to form magnesium hydroxide solid and ammonia gas. (c) Ammonium ion in aqueous solution reacts with a strong base at \(25^{\circ} \mathrm{C}\), giving aqueous ammonia and water. (c) Hydrogen sulfide gas \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) bubbled into an aqueous solution of lead(II) ions produces lead sulfide precipitate and hydrogen ions.

Hydrogen iodide gas decomposes to hydrogen gas and iodine gas: $$2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)$$ To determine the equilibrium constant of the system, identical one-liter glass bulbs are filled with \(3.20 \mathrm{~g}\) of \(\mathrm{HI}\) and maintained at a certain temperature. Each bulb is periodically opened and analyzed for iodine formation by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) $$\mathrm{I}_{2}(\mathrm{aq})+2 \mathrm{~S}_{2} \mathrm{O}_{3}{ }^{2-}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}{ }^{2-}(a q)+2 \mathrm{I}^{-}(a q)$$ It is determined that when equilibrium is reached, \(37.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) is required to titrate the iodine. What is \(K\) at the temperature of the experiment?

Isopropyl alcohol is the main ingredient in rubbing alcohol. It can decompose into acetone (the main ingredient in nail polish remover) and hydrogen gas according to the following reaction: $$\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{CO}(g)+\mathrm{H}_{2}(g)$$ At \(180^{\circ} \mathrm{C}\), the equilibrium constant for the decomposition is \(0.45\). If \(20.0 \mathrm{~mL}\) \((d=0.785 \mathrm{~g} / \mathrm{mL})\) of isopropyl alcohol is placed in a \(5.00\) -L vessel and heated to \(180^{\circ} \mathrm{C}\), what percentage remains undissociated at equilibrium?

Write a chemical equation for an equilibrium system that would lead to the following expressions (a-d) for \(K\). (a) \(K=\frac{\left(P_{\mathrm{H}_{2}}\right)^{2}\left(P_{\mathrm{O}_{2}}\right)^{3}}{\left(P_{\mathrm{SO}_{2}}\right)^{2}\left(P_{\mathrm{H}_{2} \mathrm{O}}\right)^{2}}\) (b) \(K=\frac{\left(P_{\mathrm{F}_{1}}\right)^{1 / 2}\left(P_{\mathrm{I}_{2}}\right)^{1 / 2}}{P_{\mathrm{IF}}}\) (c) \(K=\frac{\left[\mathrm{Cl}^{-}\right]^{2}}{\left(P_{\mathrm{C}_{2}}\right)\left[\mathrm{Br}^{-}\right]^{2}}\) (d) \(K=\frac{\left(P_{\mathrm{NO}}\right)^{2}\left(P_{\mathrm{H}_{3} \mathrm{O}}\right)^{4}\left[\mathrm{Cu}^{2+}\right]^{3}}{\left[\mathrm{NO}_{3}^{-}\right]^{2}\left[\mathrm{H}^{+}\right]^{8}}\)

Benzaldehyde, a flavoring agent, is obtained by the dehydrogenation of benzyl alcohol. $$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(g)+\mathrm{H}_{2}(g)$$ \(K\) for the reaction at \(250^{\circ} \mathrm{C}\) is \(0.56\). If \(1.50 \mathrm{~g}\) of benzyl alcohol is placed in a 2.0-L flask and heated to \(250^{\circ} \mathrm{C}\), (a) what is the partial pressure of the benzaldehyde when equilibrium is established? (b) how many grams of benzyl alcohol remain at equilibrium?
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