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Chapter 12: Problem 64

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\operatorname{COCl}_{2}(g) \rightleftharpoons \operatorname{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

Short Answer

Expert verified
For the given reactions, the shift in equilibrium position when the volume of the reaction container is increased is as follows: a. The equilibrium will shift to the left (towards reactants). b. The equilibrium will shift to the right (towards products). c. There will be no effect on the equilibrium position. d. The equilibrium will shift to the right (towards products). e. The equilibrium will shift to the right (towards products).

Step by step solution

01

Observe the stoichiometry of the reaction

In this reaction, 1 mole of nitrogen and 3 moles of hydrogen produce 2 moles of ammonia. The total moles of gas on the left side are 4, and on the right side, there are 2.
02

Apply Le Chatelier's principle

According to Le Chatelier's principle, increasing the volume will cause the equilibrium to shift to the side with more moles of gas. In this case, the equilibrium will shift to the left. #b - Phosphorus Pentachloride reaction#
03

Observe the stoichiometry of the reaction

In this reaction, 1 mole of phosphorus pentachloride produces 1 mole of phosphorus trichloride and 1 mole of chlorine. The total moles of gas on both sides of the reaction are equal: 1 on the left and 2 on the right.
04

Apply Le Chatelier's principle

The principle states that the equilibrium will shift to the side with more moles of gas when the volume is increased. In this case, the equilibrium will shift to the right. #c - Hydrogen and Fluorine reaction#
05

Observe the stoichiometry of the reaction

In this reaction, 1 mole of hydrogen and 1 mole of fluorine produce 2 moles of hydrogen fluoride. The total moles of gas on both sides of the reaction are equal: 2 on the left and 2 on the right.
06

Apply Le Chatelier's principle

Since the total moles of gas on both sides of the reaction are equal, increasing the volume will have no effect on the equilibrium position. #d - Carbonyl Chloride reaction#
07

Observe the stoichiometry of the reaction

In this reaction, 1 mole of carbonyl chloride produces 1 mole of carbon monoxide and 1 mole of chlorine. The total moles of gas on both sides of the reaction are equal: 1 on the left and 2 on the right.
08

Apply Le Chatelier's principle

According to Le Chatelier's principle, increasing the volume will cause the equilibrium to shift to the side with more moles of gas. In this case, the equilibrium will shift to the right. #e - Calcium Carbonate reaction#
09

Observe the stoichiometry of the reaction

In this reaction, 1 mole of calcium carbonate produces 1 mole of calcium oxide and 1 mole of carbon dioxide. The left side has no moles of gas (only solid), and the right side has 1 mole of gas (CO2).
10

Apply Le Chatelier's principle

According to the principle, increasing the volume will cause the equilibrium to shift to the side with more moles of gas. In this case, the equilibrium will shift to the right.

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Most popular questions from this chapter

The following equilibrium pressures at a certain temperature were observed for the reaction $$\begin{aligned}2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\P_{\mathrm{NO}_{2}} &=0.55 \mathrm{atm} \\\P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{atm} \\\P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{atm}\end{aligned}$$. Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.

For a typical equilibrium problem, the value of \(K\) and the initial reaction conditions are given for a specific reaction, and you are asked to calculate the equilibrium concentrations. Many of these calculations involve solving a quadratic or cubic equation. What can you do to avoid solving a quadratic or cubic equation and still come up with reasonable equilibrium concentrations?

Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g),\) consider two possibilities: (a) you mix 0.5 mole of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix 1.5 moles of \(\mathrm{H}_{2}\) and 0.5 mole of \(\mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Nitrogen gas \(\left(\mathrm{N}_{2}\right)\) reacts with hydrogen gas \(\left(\mathrm{H}_{2}\right)\) to form ammonia \(\left(\mathrm{NH}_{3}\right) .\) At \(200^{\circ} \mathrm{C}\) in a closed container, 1.00 atm of nitrogen gas is mixed with 2.00 atm of hydrogen gas. At equilibrium,the total pressure is 2.00 atm. Calculate the partial pressure of hydrogen gas at equilibrium, and calculate the \(K_{\mathrm{p}}\) value for this reaction.
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