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Chapter 12: Problem 68

Hydrogen, bromine, and \(\mathrm{HBr}\) in the gas phase are in equilibrium in a container of fixed volume. \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g}) \quad \Delta_{r} H^{\circ}=-103.7 \mathrm{~kJ} / \mathrm{mol}\) How will each of these changes affect the indicated quantities? Write "increase," "decrease," or "no change." \begin{tabular}{l} \hline Change & {\(\left[\mathrm{Br}_{2}\right]\)} & {\([\mathrm{HBr}]\)} & \(K_{c}\) & \(K_{\mathrm{p}}\) \\ \hline Some \(\mathrm{H}_{2}\) is added to the \\ container. \\ The temperature of the gases \\ in the container is increased. \\ The pressure of \(\mathrm{HBr}\) is \\ increased. \end{tabular}

Short Answer

Expert verified
Adding H2: [Br2] decreases, [HBr] increases, Kc no change, Kp no change. Increasing temperature: [Br2] increases, [HBr] decreases, Kc decreases, Kp decreases. Increasing HBr pressure: [Br2] increases, [HBr] decreases, Kc no change, Kp no change.

Step by step solution

01

Analyze the effect of adding H2

According to Le Chatelier's principle, if a chemical equilibrium is disturbed by adding more reactants, the system will shift to counteract the change. Adding more \( \text{H}_2 \) will shift the equilibrium to the right to produce more \( \text{HBr} \), trying to consume the excess \( \text{H}_2 \). Hence, \( \left[ \text{Br}_2 \right] \) decreases, \( \left[ \text{HBr} \right] \) increases, and there is no change in \( K_c \) and \( K_p \) as they only depend on temperature.
02

Analyze the effect of increasing the temperature

The reaction is exothermic (\( \Delta_r H^\circ = -103.7 \text{ kJ/mol} \)), meaning heat is released when the reaction proceeds to the right. Increasing temperature will favor the endothermic reverse reaction, shifting the equilibrium to the left. Therefore, \( \left[ \text{Br}_2 \right] \) increases, \( \left[ \text{HBr} \right] \) decreases, and both \( K_c \) and \( K_p \) decrease because they are affected by changes in temperature for exothermic processes.
03

Analyze the effect of increasing the pressure of HBr

Increasing the pressure of \( \text{HBr} \) specifically will push the equilibrium towards the left side to reduce the concentration of \( \text{HBr} \). As a result, \( \left[ \text{Br}_2 \right] \) increases, \( \left[ \text{HBr} \right] \) decreases. \( K_c \) and \( K_p \) remain unchanged since they are dependent only on temperature.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium refers to the state in a chemical reaction where both the forward and reverse reactions occur at the same rate. At this point, the concentrations of reactants and products remain constant over time. This doesn't mean they are equal, but rather that their rates match, creating a balance in the system.

Le Chatelier's Principle helps us predict how changes in conditions will shift the equilibrium. If a stress, such as a change in concentration, temperature, or pressure, is applied to a system at equilibrium, the system will adjust to counteract the stress.
  • Adding a reactant will typically shift the equilibrium towards the products to use the added substance.
  • Removing a reactant or adding a product will shift the balance towards the reactants.
In this way, the system naturally tries to restore equilibrium. A crucial aspect here is understanding that the equilibrium constant ( K c or Kp) itself does not change with alterations in concentration or pressure. It is primarily sensitive to changes in temperature.
Exothermic Reactions
An exothermic reaction is a type of chemical reaction that releases heat into the surroundings. In the context of chemical equilibrium, the presence of heat in exothermic reactions plays a vital role in determining how the equilibrium position will shift when temperature changes.

For an exothermic reaction, such as the formation of HBr gas from hydrogen and bromine gas, increasing the temperature generally makes the system favor the reverse reaction. This is because the system absorbs some of this added heat by shifting towards the side that consumes heat - the reactants side in this case.
  • This shift means the concentration of products will decrease while reactants will increase.
  • Conversely, lowering the temperature will favor product formation, striving to release heat to compensate for the temperature drop.
Thus, understanding how temperature interacts with exothermic reactions can help predict the direction in which equilibrium will shift.
Equilibrium Constant
The equilibrium constant ( K ) is a number that expresses the relationship between the concentrations of products and reactants at equilibrium. For a given chemical reaction, it can be represented in two common forms: Kc, which is based on molarity, and Kp, based on partial pressures for gaseous reactions.

Kc and Kp are calculated using the balanced chemical equation, taking the concentrations or partial pressures raised to the power of their coefficients. This allows us to quantify the exact position of equilibrium quantitatively. While the concentrations of individual reactants and products might change with adjustments in physical conditions like pressure and concentration, the equilibrium constant itself remains unaffected by these changes.
  • The only variable that impacts the values of Kc and Kp is temperature.
  • For exothermic reactions, an increase in temperature typically results in a decrease in both Kc and Kp.
  • On the other hand, lowering the temperature would generally increase Kc and Kp, reflecting more formation of products.
Understanding the equilibrium constant provides a clear insight into how far a reaction will proceed to reach equilibrium under specific conditions.

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Most popular questions from this chapter

Write equilibrium constant expressions, in terms of reactant and product concentrations, for each of these reactions. $$ \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \quad K_{\mathrm{c}}=1.0 \times 10^{-14} $$ \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) $$ \begin{array}{c} K_{\mathrm{c}}=1.8 \times 10^{-5} \\ \mathrm{~N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \end{array} $$ Assume that all gases and solutes have initial concentrations of \(1.0 \mathrm{~mol} / \mathrm{L}\). Then let the first reactant in each reaction change its concentration by \(-x\). (a) Using the reaction table (ICE table) approach, write equilibrium constant expressions in terms of the unknown variable \(x\) for each reaction. (b) Which of these expressions yield quadratic equations? (c) How would you go about solving the others for \(x ?\)

Carbon tetrachloride can be produced by this reaction: $$ \mathrm{CS}_{2}(\mathrm{~g})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{S}_{2} \mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{CCl}_{4}(\mathrm{~g}) $$ Suppose \(1.2 \mathrm{~mol} \mathrm{CS}_{2}\) and \(3.6 \mathrm{~mol} \mathrm{Cl}_{2}\) are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains \(0.90 \mathrm{~mol} \mathrm{CCl}_{4} \cdot\) Calculate \(K_{\mathrm{c}}\).

Write the expression for \(K_{\mathrm{c}}\) for each reaction. (a) \(\mathrm{PCl}_{5}(\mathrm{~s}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (b) \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}(\mathrm{aq})+4 \mathrm{Cl}^{-}(\mathrm{aq}) \rightleftharpoons\) \(\mathrm{CoCl}_{4}^{2-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})\) (d) \(2 \mathrm{~F}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{OF}_{2}(\mathrm{~g})+2 \mathrm{HF}(\mathrm{g})\)

Write equilibrium constant expressions for these reactions. For gases, use either pressures or concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{~g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{~g})\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{~s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)

Two molecules of A react to form one molecule of \(\mathrm{B},\) as in the reaction $$ 2 \mathrm{~A}(\mathrm{~g}) \rightleftharpoons \mathrm{B}(\mathrm{g}) $$ Three experiments are done at different temperatures and equilibrium concentrations are measured. For each experiment, calculate the equilibrium constant, \(K_{\mathrm{c}^{*}}\) (a) \([\mathrm{A}]=0.74 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.74 \mathrm{~mol} / \mathrm{L}\) $$ \begin{array}{l} \text { (b) }[\mathrm{A}]=2.0 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=2.0 \mathrm{~mol} / \mathrm{L} \\ \text { (c) }[\mathrm{A}]=0.01 \mathrm{~mol} / \mathrm{L},[\mathrm{B}]=0.01 \mathrm{~mol} / \mathrm{L} \end{array} $$ What can you conclude about this statement: "If the concentrations of reactants and products are equal, then the equilibrium constant is always \(1.0 . "\)
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