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The four substances \(\mathrm{HCl}, \mathrm{I}_{2}, \mathrm{HI}\), and \(\mathrm{Cl}_{2}\) are mixed in a reaction vessel and allowed to reach equilibrium in the reaction \(2 \mathrm{HCl}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g})\). Certain changes (which are specified in the first column in the following table) are then made to this mixture. Considering each change separately, state the effec (increase, decrease, or no change) that the change has on the original equilibrium value of the quantity in the second column (or \(K\), if that is specified). The temperature and volume are constant. Change Quantity (a) add \(\mathrm{HCl} \quad\) amount of HI (b) add \(\mathrm{I}_{2} \quad\) amount of \(\mathrm{Cl}_{2}\) (c) remove HI amount of \(\mathrm{Cl}_{2}\) (d) remove \(\mathrm{Cl}_{2} \quad\) amount of \(\mathrm{HCl}\) (e) add \(\mathrm{HCl} \quad K\) (f) remove \(\mathrm{HCl}\) amount of \(\mathrm{I}_{2}\) (g) add \(\mathrm{I}_{2} \quad \mathrm{~K}\)

Short Answer

Expert verified
The amount of HI will increase for (a), the amount of Cl2 will increase for (b) and (c), the amount of HCl will decrease for (d), there will be no change in K for (e) and (g), and the amount of I2 will decrease for (f).

Step by step solution

01

Analyzing the Effect of Adding HCl

By adding more HCl, the reaction will shift to the right according to Le Chatelier’s Principle in order to counteract the change. This means the concentration of HI will increase as HCl is a reactant.
02

Analyzing the Effect of Adding I2

Adding more I2 will also cause the reaction to shift to the right. As a result, the concentration of Cl2 will increase because iodine (I2) is a reactant.
03

Analyzing the Effect of Removing HI

Removing HI from the system will cause a shift to the right as the reaction seeks to replace the removed HI. This will lead to an increase in the concentration of Cl2.
04

Analyzing the Effect of Removing Cl2

Removing Cl2 will drive the equilibrium to the left to produce more Cl2 from the reactants. Therefore, the amount of HCl will decrease.
05

Evaluating the Impact on Equilibrium Constant K after Adding HCl

The equilibrium constant K is dependent on temperature only. Since temperature is held constant, adding HCl will have no effect on the value of K.
06

Analyzing the Effect of Removing HCl

Removing HCl will cause the position of equilibrium to shift to the left to produce more HCl. This will result in a decrease in the amount of I2, since it's a product of the reverse reaction.
07

Evaluating the Impact on Equilibrium Constant K after Adding I2

Similar to the case with HCl, adding I2 does not affect the equilibrium constant K since the temperature remains constant.

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Key Concepts

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

Le Chatelier’s Principle
Understanding Le Chatelier’s Principle will enable students to predict how a chemical system at equilibrium responds to changes. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In simpler terms, it's like a seesaw balancing act: if you add more weight to one side, the other side will react to try and level it out again.

When additional reactants are introduced to the system, as in the exercise's scenarios (a) and (b), equilibrium shifts to the right, meaning that the system will favor the production of products to re-establish equilibrium. Conversely, removing a product, as in scenarios (c) and (d), will shift the equilibrium to the left, favoring the formation of reactants. It’s important to emphasize that Le Chatelier’s Principle offers insight into the direction of the shift, not the extent of the change or the new concentrations of the substances involved.
Equilibrium Constant K
The equilibrium constant, symbolized as K, is a number that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction at a constant temperature. It's essential for students to understand that K remains unchanged unless the temperature changes. This distinction is crucial in scenarios (e) and (g) from the exercise.

Adding or removing a reactant or product does not alter the value of K. For example, in scenario (e), when more HCl is added, the system counteracts the increase by shifting towards the products, but K remains the same because the temperature does not change. Students should remember the formula for the equilibrium constant and how it is unaffected by changes in concentrations of reactants or products. The formula for a general reaction \( aA + bB \rightleftharpoons cC + dD \) is \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) where the brackets indicate concentration, and the lowercase letters represent the stoichiometric coefficients from the balanced chemical equation.
Equilibrium Shift
An 'equilibrium shift' refers to the movement of a chemical reaction's position of equilibrium in response to a change in concentration, temperature, pressure, or volume. Students should understand that an equilibrium shift is the system's way of regaining balance. During this shift, the concentrations of reactants and products change to reach a new equilibrium state. However, it's vital to note that these changes do not affect the equilibrium constant K if temperature remains constant, reinforcing the concept highlighted in the section above.

In the context of the exercise, when a change occurs such as the addition or removal of a substance, the equilibrium will shift to increase or decrease the amounts of substances opposite to the change—always aiming to minimize the disturbance. This concept is exemplified in scenarios (b) and (c) where adding more \( I_2 \) or removing HI respectively, leads to an increased amount of \( Cl_2 \) as the reaction seeks to re-establish its equilibrium position.

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Most popular questions from this chapter

if \(Q=1.0\) for the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants or will it be at equilibrium?

The four gases \(\mathrm{NH}_{3}, \mathrm{O}_{2}, \mathrm{NO}\), and \(\mathrm{H}_{2} \mathrm{O}\) are mixed in a reaction vessel and allowed to reach equilibrium in the reaction \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). Certain changes (see the following table) are then made to this mixture. Considering each change separately, state the effect (increase, decrease, or no change) that the change has on the original equilibrium values of the quantity in the second column (or \(K\), if that is specified). The temperature and volume are constant. \(\begin{array}{ll}\text { Change } & \text { Quantity } \\ \text { (a) add } \mathrm{NO} & \text { amount of } \mathrm{H}_{2} \mathrm{O} \\ \text { (b) add } \mathrm{NO} & \text { amount of } \mathrm{O}_{2} \\ \text { (c) remove } \mathrm{H}_{2} \mathrm{O} & \text { amount of } \mathrm{NO} \\ \text { (d) remove } \mathrm{O}_{2} & \text { amount of } \mathrm{NH}_{3} \\ \text { (e) add } \mathrm{NH}_{3} & \mathrm{~K} \\ \text { (f) remove } \mathrm{NO} & \text { amount of } \mathrm{NH}_{3} \\ \text { (g) add } \mathrm{NH}_{3} & \text { amount of } \mathrm{O}_{2}\end{array}\)

In a gas-phase equilibrium mixture of \(\mathrm{SbCl}_{5}, \mathrm{SbCl}_{3}\), and \(\mathrm{Cl}_{2}\) at \(500 . \mathrm{K}, P_{\mathrm{SbCl}_{5}}=0.072\) bar and \(P_{\mathrm{SbCl}_{3}}=5.02 \mathrm{mbar}\). Calculate the equilibrium partial pressure of \(\mathrm{Cl}_{2}\), given that \(K=3.5 \times 10^{-4}\) for the reaction \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\).

(a) Calculate the reaction Gibbs free energy of \(\mathrm{N}_{2}\) (g) \(+\) \(3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) when the partial pressures of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are \(4.2\) bar, \(1.8\) bar, and 21 bar, respectively, and the temperature is 400 . K. For this reaction, \(K=41\) at 400 . K. (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

Let \(\alpha\) be the fraction of \(\mathrm{PCl}_{5}\) molecules that have decomposed to \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) in the reaction \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\) \(\mathrm{Cl}_{2}(\mathrm{~g})\) in a constant- volume container; then the amount of \(\mathrm{PCl}_{5}\) at equilibrium is \(n(1-\alpha)\), where \(n\) is the amount present initially. Derive an equation for \(K\) in terms of \(\alpha\) and the total pressure \(P\), and solve it for \(\alpha\) in terms of \(P\). Calculate the fraction decomposed at \(556 \mathrm{~K}\), at which temperature \(K=4.96\), and the total pressure is (a) \(0.50\) bar; (b) \(1.00\) bar.

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