State whether reactants or products will be favored by compression in each of the following equilibria. If no change occurs, explain why that is so. (a) \(2 \mathrm{O}_{3}(\mathrm{~g})=3 \mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{g})\) (c) \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g})=4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(2 \mathrm{HD}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2}(\mathrm{~g})\) (e) \(\mathrm{Cl}_{2}(\mathrm{~g})=2 \mathrm{Cl}(\mathrm{g})\)

Short Answer

Expert verified
Compression will favor the reactants for (a) and (e), have no effect on (b) and (d), and favor the reactants for (c).

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. In the context of compression, this means that increasing the pressure will shift the equilibrium towards the side with fewer moles of gas.
02

Analyze Reaction (a)

For the reaction 2 O3(g) = 3 O2(g), compression will favor the reactants because there are fewer moles of gas on the reactants side (2 moles) compared to the products side (3 moles).
03

Analyze Reaction (b)

In the reaction H2O(g) + C(s) ⇌ H2(g) + CO(g), compression will have no effect on the position of equilibrium. Solids don't contribute to the pressure in a closed system, so the mole ratio of gases is 1:2 before and after the reaction, meaning the number of moles of gases remains the same on both sides.
04

Analyze Reaction (c)

For 4 NH3(g) + 5 O2(g) = 4 NO(g) + 6 H2O(g), compression will favor the reactants because there are more moles of gas on the products side (10 moles) than on the reactants side (9 moles).
05

Analyze Reaction (d)

The reaction 2 HD(g) ⇌ H2(g) + D2(g) has the same number of moles of gas on both sides of the equation (2 moles). Therefore, compression will not favor either side, and no change will occur in the position of equilibrium.
06

Analyze Reaction (e)

For the dissociation of Cl2(g) to 2 Cl(g), compression will favor the reactants because the reactants have fewer moles of gas (1 mole) compared to the products (2 moles).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Equilibrium
Understanding chemical equilibrium is fundamental to grasping how reactions behave under various conditions. Equilibrium occurs in a closed system when the rates of the forward and reverse reactions are equal, leading to a constant ratio of product and reactant concentrations.

This balance does not mean the reactions have stopped; rather, they continue to occur simultaneously but with no net change in concentration of reactants and products. It's a dynamic state where chemical substances are synthesized and broken down at an equal pace. When equilibrium is achieved, it can be represented by a balanced equation where the amount of each substance remains steady over time.

For students who might be new to this concept, imagine a crowded room where an equal number of people are entering and leaving at the same time. The total number of people in the room stays the same, much like the molecules in a system at equilibrium.
Reaction Compression Effects
Reaction compression effects refer to the changes that occur in a reaction at equilibrium when the pressure is increased, typically due to a decrease in volume. According to Le Chatelier's Principle, an increase in pressure will cause the equilibrium to shift in the direction that will reduce pressure.

One key fact to remember is that this change is associated with the number of moles of gas; when there are more moles of gas on one side of a balanced equation, and pressure is increased, the system will adjust by favoring the side with fewer moles of gas. For instance, a balloon squeezed under pressure will shrink to minimize its volume and reduce pressure, and a similar adjustment happens on the molecular level in reactions occurring in a compressed system.
Equilibrium Shift
An equilibrium shift happens when a change in conditions—such as temperature, pressure, concentration, or volume—forces a system at equilibrium to adjust in order to restore balance. This shift can favor either the production of reactants or products, depending upon the nature of the change imposed.

For example, in the reaction mentioned in the exercise, (a) where compression causes the equilibrium to shift towards the reactants due to having fewer moles of gas, it is analogous to seeing the scales tip in the direction that restores balance—think of it as nature's way of keeping things steady. Remember, equilibrium will not shift if the change in condition does not affect the balance between reactants and products, such as in cases where the number of gas moles remains unchanged.

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Most popular questions from this chapter

The temperature dependence of the equilibrium constant of the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\), which makes an important contribution to atmospheric nitrogen oxides, can be expressed as \(\ln K=2.5-\) \((21700 \mathrm{~K}) \cdot T^{-1}\). What is the standard enthalpy of the forward reaction?

\( \mathrm{~A} 3.00-\mathrm{L}\) reaction vessel is filled with \(0.150 \mathrm{~mol}\) \(\mathrm{CO}, 0.0900 \mathrm{~mol} \mathrm{H}_{2}\), and \(0.180 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\). Equilibrium is reached in the presence of a zinc oxidechromium(III) oxide catalyst; and at \(300^{\circ} \mathrm{C}, K_{c}=\) \(1.1 \times 10^{-2}\) for the reaction, \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\). (a) As the reaction approaches equilibrium, will the molar concentration of \(\mathrm{CH}_{3} \mathrm{OH}\) increase, decrease, or remain unchanged? (b) What is the equilibrium composition of the mixture?

If \(Q=1.0 \times 10^{50}\) for the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

State what happens to the concentration of the indicated substance when each of the following equilibrium systems is compressed. (a) \(\mathrm{NO}_{2}\) (g) in \(2 \mathrm{~Pb}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{~s})=\) \(2 \mathrm{PbO}(\mathrm{s})+4 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) (b) \(\mathrm{NO}(\mathrm{g})\) in \(3 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I})=\) \(2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{g})\) (c) \(\mathrm{HI}(\mathrm{g})\) in \(2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~s})=2 \mathrm{Hl}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (d) \(\mathrm{SO}_{2}\) (g) in \(2 \mathrm{SO}_{2}\) (g) \(+\mathrm{O}_{2}\) (g) \(\rightleftharpoons 2 \mathrm{SO}_{3}\) (g) (c) \(\mathrm{NO}_{2}(\mathrm{~g})\) in \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\)

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