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Chapter 12: Problem 63

Suppose the reaction system $$\mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional UO \(_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

Short Answer

Expert verified
The effect of the given changes on the equilibrium position are as follows: a. Adding additional UO\(_{2}(s)\) will have no effect on the equilibrium position because the concentration of a solid is constant and unchanged. b. When HF(g) reacts with the glass reaction vessel, the equilibrium will shift to the left to counteract the decrease in HF(g) concentration. c. The removal of water vapor will cause the equilibrium to shift to the right to compensate for the decrease in H\(_{2}\)O(g) concentration.

Step by step solution

01

Write the equilibrium reaction

We are given the following equilibrium reaction system: \[ \mathrm{UO}_{2}(s) + 4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g) + 2 \mathrm{H}_{2} \mathrm{O}(g) \] Now we can analyze the effect of the given changes on this equilibrium system. #Step 2: Analyze the effect of adding additional UO\(_{2}(s)\).#
02

Analyze the effect of adding additional UO\(_{2}(s)\)

Since UO\(_{2}(s)\) is a solid, its concentration is constant and unaffected by the amount present in the system. Adding more UO\(_{2}(s)\) does not change its active concentration in the reaction. Therefore, there is no effect on the equilibrium, and the equilibrium position does not shift. #Step 3: Analyze the effect of the reaction between HF(g) and the glass reaction vessel.#
03

Analyze the effect of the reaction between HF(g) and the glass reaction vessel

The reaction between HF(g) and the glass reaction vessel leads to the consumption of HF(g). This results in a decrease in the concentration of HF(g). According to Le Châtelier's principle, the equilibrium will shift in the direction that counteracts this change, which is the direction where HF(g) is formed. In our case, that direction is towards the reactants. Therefore, the equilibrium will shift to the left. #Step 4: Analyze the effect of removing water vapor.#
04

Analyze the effect of removing water vapor

When the water vapor is removed, the concentration of H\(_{2}\)O(g) decreases. According to Le Châtelier's principle, the system will respond by shifting the equilibrium to counteract this change, which means shifting the equilibrium in the direction where H\(_{2}\)O(g) is formed. In our case, that direction is towards the products. Therefore, the equilibrium will shift to the right.

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Most popular questions from this chapter

In a given experiment, 5.2 moles of pure NOCl was placed in an otherwise empty \(2.0-\mathrm{L}\) container. Equilibrium was established by the following reaction: $$2 \operatorname{Nocl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \quad K=1.6 \times 10^{-5}$$, a. Using numerical values for the concentrations in the Initial row and expressions containing the variable \(x\) in both the Change and Equilibrium rows, complete the following table summarizing what happens as this reaction reaches equilibrium. Let \(x=\) the concentration of \(\mathrm{Cl}_{2}\) that s present at equilibrium. b. Calculate the equilibrium concentrations for all species.

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Ammonia is produced by the Haber process, in which nitrogen and hydrogen are reacted directly using an iron mesh impregnated with oxides as a catalyst. For the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$.equilibrium constants \(\mathbf{r}_{\mathbf{p}}\).\(300^{\circ} \mathrm{C}, \quad 4.34 \times 10^{-3}\) \(500^{\circ} \mathrm{C}, \quad 1.45 \times 10^{-5}\) \(600^{\circ} \mathrm{C}, \quad 2.25 \times 10^{-6}\) Is the reaction exothermic or endothermic?

Consider the following reaction:$$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)$$.Amounts of \(\mathrm{H}_{2} \mathrm{O}, \mathrm{CO}, \mathrm{H}_{2},\) and \(\mathrm{CO}_{2}\) are put into a flask so that the composition corresponds to an equilibrium position. If the CO placed in the flask is labeled with radioactive \(^{14} \mathrm{C},\) will \(^{14} \mathrm{C}\) be found only in CO molecules for an indefinite period of time? Explain.
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