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Consider the equilibrium: $$ 2 \mathrm{I}(g) \rightleftarrows \mathrm{I}_{2}(g) $$ What would be the effect on the position of equilibrium of (a) increasing the total pressure on the system by decreasing its volume, (b) adding \(\mathrm{I}_{2}\) to the reaction mixture, and (c) decreasing the temperature?

Short Answer

Expert verified
a) Right; b) Left; c) Right (assuming exothermic).

Step by step solution

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01

Analyze Effect of Pressure Increase

When the total pressure is increased by decreasing the volume, according to Le Chatelier's Principle, the equilibrium will shift towards the side with fewer gas molecules to counteract the change. In the given reaction, \[ 2 \mathrm{I}(g) \rightleftarrows \mathrm{I}_{2}(g) \]there are more moles of gas on the left (2 moles of \(\mathrm{I}\)) than on the right (1 mole of \(\mathrm{I}_{2}\)). Thus, increasing pressure will shift the equilibrium to the right.
02

Effect of Adding I2

When \(\mathrm{I}_{2}\) is added to the mixture, the system will counteract the change by consuming some of the additional \(\mathrm{I}_{2}\). According to Le Chatelier's Principle, this will shift the equilibrium to the left, favoring the formation of \(\mathrm{I}\) gas.
03

Effect of Decreasing Temperature

The effect of changing temperature on equilibrium depends on whether the reaction is exothermic or endothermic. If the reaction is exothermic (which we assume here as a common condition for dimerization reactions like this), decreasing the temperature will shift the equilibrium towards the products, \(\mathrm{I}_{2}\), to release heat. Thus, the equilibrium position will shift to the right.

Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle helps us understand how a chemical equilibrium reacts to changes in conditions. It states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift to counteract the imposed change and re-establish equilibrium. This principle is an essential tool for predicting the direction of the equilibrium shift under different experimental conditions.
  • For example, adding more of a reactant will typically shift the equilibrium toward the products.
  • Conversely, removing a product will also shift the balance toward the products to fill the void left.
Understanding this principle allows chemists to control the outcome of reactions by manipulating conditions to favor the production of desired substances.
Effect of Pressure on Equilibrium
The effect of pressure on chemical equilibrium is significant, especially for reactions involving gases. According to Le Chatelier's Principle, an increase in pressure by reducing volume causes the equilibrium to shift toward the side with fewer gas molecules.
In the reaction \[2 \mathrm{I}(g) \rightleftarrows \mathrm{I}_{2}(g)\]there are more moles of gaseous iodine on the left than iodine molecules on the right. Hence, increasing the pressure will push the equilibrium towards the side with fewer moles, which is the right side in this case, favoring the formation of iodine molecules \(\mathrm{I}_{2}\).
  • This shift reduces the total number of gas molecules, thus opposing the pressure increase.
  • Decreasing the pressure would result in the equilibrium shifting back to the left, favoring more gaseous iodine formation.
Effect of Temperature on Equilibrium
Temperature changes can have a profound effect on chemical equilibria. According to Le Chatelier's Principle, the equilibrium will shift in a direction that offsets the change in temperature. The reaction type, whether endothermic or exothermic, plays a crucial role in determining this shift.
  • In exothermic reactions, such as the dimerization of iodide, decreasing temperature favors the production of more products (\(\mathrm{I}_{2}(g)\)).
  • This is because the system releases heat, shifting the equilibrium to the right to produce more heat in response to the temperature drop.
If the reaction were endothermic, decreasing the temperature would result in shifting to the left, favoring the reactants.
Equilibrium Shift
An equilibrium shift occurs when a change, such as pressure, concentration, or temperature, is applied to a system in equilibrium. This shift attempts to counterbalance the change and re-establish equilibrium. Knowing the direction of these shifts helps predict the outcome of changes in reaction conditions.
In our example of \[2 \mathrm{I}(g) \rightleftarrows \mathrm{I}_{2}(g)\]- Adding \(\mathrm{I}_{2}\) increases the concentration of products, causing a shift to the left.- This makes more \(\mathrm{I}(g)\), as the system tries to reduce this added \(\mathrm{I}_{2}\) by converting it back into gases.
  • This mechanism underscores the dynamic nature of chemical equilibria, constantly adjusting to maintain balance.
Reaction Dynamics
Reaction dynamics focus on the factors that affect the speed and position of a reaction at equilibrium. These dynamics encompass temperature, pressure, and concentration changes, all of which are interrelated in influencing equilibrium.
- When altering conditions, an equilibrium's response can yield insights into the interaction of reactants and products. - Reaction dynamics also enable us to tailor conditions to favor desired equilibrium outcomes in industrial and laboratory processes.
  • For example, increasing pressure or decreasing the volume in gas-phase reactions can speed up the formation of products by shifting equilibrium optimally.
  • Similarly, temperature adjustments can be critical in maximizing yield for industrial chemical syntheses.
Recognizing how dynamics play into equilibrium shifts is vital for optimizing chemical processes.

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Most popular questions from this chapter

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant, and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftarrows 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftarrows \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftarrows \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftarrows \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftarrows 2 \mathrm{~B}(g)\)

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction: $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ is 4.2 at \(1650^{\circ} \mathrm{C}\). Initially \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.80 \mathrm{~mol}\) \(\mathrm{CO}_{2}\) are injected into a 5.0-L flask. Calculate the concentration of each species at equilibrium.

Write the equation for the reaction that corresponds to each of the following reaction quotients: (a) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{H}_{2}\right]^{2}\left[\mathrm{~S}_{2}\right]}{\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}\) (b) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{NO}_{2}\right]^{2}\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{NClO}_{2}\right]^{2}}\) (c) \(Q_{\mathrm{c}}=\frac{\left[\mathrm{HgI}_{4}^{2-}\right]}{\left[\mathrm{Hg}^{2+}\right]\left[\mathrm{I}^{-}\right]^{4}}\) (d) \(Q_{\mathrm{c}}=\frac{[\mathrm{NO}]^{2}\left[\mathrm{Br}_{2}\right]}{[\mathrm{NOBr}]^{2}}\)

The formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is an intermediate step in the manufacture of sulfuric acid, and it is also responsible for the acid rain phenomenon. The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{SO}_{3}(g)$$ is 0.13 at \(830^{\circ} \mathrm{C}\). In one experiment, \(2.00 \mathrm{~mol} \mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol} \mathrm{O}_{2}\) were initially present in a flask. What must the total pressure at equilibrium be to have an 80.0 percent yield of \(\mathrm{SO}_{3} ?\)

Photosynthesis can be represented by: $$\begin{aligned} 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \rightleftarrows \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \\ \Delta H^{\circ} &=2801 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ Explain how the equilibrium would be affected by the following changes: (a) partial pressure of \(\mathrm{CO}_{2}\) is increased, (b) \(\mathrm{O}_{2}\) is removed from the mixture, (c) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose) is removed from the mixture, (d) more water is added, (e) a catalyst is added, (f) temperature is decreased.

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