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The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. \(P_{\mathrm{H}_{2} \mathrm{O}}=1.00 \mathrm{~atm}, P_{\mathrm{Cl}_{2} \mathrm{O}}=1.00 \mathrm{~atm}, P_{\mathrm{HOCl}}=1.00 \mathrm{~atm}\) b. \(P_{\mathrm{H}_{2} \mathrm{O}}=200 .\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=49.8\) torr, \(P_{\mathrm{HOCl}}=21.0\) torr c. \(P_{\mathrm{H}_{2} \mathrm{O}}=296\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=15.0\) torr, \(P_{\mathrm{HOCl}}=20.0\) torr

Short Answer

Expert verified
For case a: The system is not at equilibrium and will shift to the left (favoring reactants). For case b: The system is not at equilibrium and will shift to the left (favoring reactants). For case c: The system is not at equilibrium and will shift to the right (favoring products).

Step by step solution

01

Calculate the reaction quotient (Q)

The reaction quotient (Q) can be calculated using the partial pressures of the reactants and products: \[Q = \frac{P_{\mathrm{HOCl}}^2}{P_{\mathrm{H}_{2}\mathrm{O}} \cdot P_{\mathrm{Cl}_2\mathrm{O}}}\] Now let's evaluate Q for each case.
02

Case a:

Calculate the Q value using the given partial pressures: \[Q = \frac{(1.00)^2}{(1.00)(1.00)} = 1\]
03

Case b:

Before calculating Q value, we need to convert torr to atm: \[1 \mathrm{~atm} = 760 \mathrm{~torr}\] \[P_{\mathrm{H}_{2}\mathrm{O}} = \frac{200}{760} \mathrm{~atm}\] \[P_{\mathrm{Cl}_{2}\mathrm{O}} = \frac{49.8}{760} \mathrm{~atm}\] \[P_{\mathrm{HOCl}} = \frac{21.0}{760} \mathrm{~atm}\] Now, calculate the Q value: \[Q = \frac{(\frac{21.0}{760})^2}{(\frac{200}{760})(\frac{49.8}{760})} \approx 1.966\]
04

Case c:

Convert torr to atm and then calculate the Q value: \[P_{\mathrm{H}_{2}\mathrm{O}} = 296/760 \mathrm{~atm}\] \[P_{\mathrm{Cl}_{2}\mathrm{O}} = 15.0/760 \mathrm{~atm}\] \[P_{\mathrm{HOCl}} = 20.0/760 \mathrm{~atm}\] \[Q = \frac{(20.0/760)^2}{(296/760)(15.0/760)} \approx 0.0172\]
05

Determine the state of the system and shifts

Compare Q with K to see if the system is at equilibrium. If Q > K, the reaction will shift to the left to reach equilibrium. If Q < K, the reaction will shift to the right to reach equilibrium. If Q = K, the system is at equilibrium. K = 0.0900
06

Case a:

Q = 1 > K = 0.0900, so the reaction will shift to the left (favoring the reactants) to reach equilibrium.
07

Case b:

Q ≈ 1.966 > K = 0.0900, thus the reaction will shift to the left (favoring the reactants) to reach equilibrium.
08

Case c:

Q ≈ 0.0172 < K = 0.0900, so the reaction will shift to the right (favoring the products) to reach equilibrium.

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Key Concepts

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

Chemical Equilibrium
Understanding the concept of chemical equilibrium is fundamental in the study of chemical reactions. It occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to a balance in the concentrations of reactants and products. This balance does not imply that the reactants and products are at equal concentrations, but rather that their ratio remains constant over time.

Once achieved, the system is said to be at equilibrium, represented by the equilibrium constant (K). The value of K is specific to each chemical reaction and is influenced by temperature. It provides a way of determining the direction in which a reaction mixture will shift to reach equilibrium when the system is not currently at that state. It's crucial to realize that equilibrium describes a dynamic process where reactions continue to occur, but with no net change in the concentrations of reactants and products.
Reaction Quotient
The reaction quotient, denoted as Q, plays a pivotal role in predicting the direction of a reaction's shift towards equilibrium. It is calculated in the same way as the equilibrium constant, but with the current concentrations or partial pressures of the reactants and products, rather than the concentrations at equilibrium.

Comparing Q to the equilibrium constant (K) indicates which way the reaction needs to shift to reach equilibrium:
  • If Q < K, the reaction will proceed in the forward direction, increasing the production of products.
  • If Q > K, the reaction will proceed in the reverse direction, increasing the concentration of reactants.
  • If Q = K, the reaction is at equilibrium, and no shift is needed.
By calculating Q for various sets of conditions, we can predict how a system not at equilibrium will behave to establish that balance.
Partial Pressure
Partial pressure is a term often used when discussing gases involved in chemical reactions. It is the pressure that a particular gas in a mixture of gases would exert if it alone occupied the entire volume. In the context of reaction quotients and equilibrium constants, the partial pressure of a gaseous substance plays a crucial role in calculating Q and K when dealing with gas-phase reactions.

It's also important to understand how to convert measurements of pressure, such as from torr to atm, as seen in the provided exercise. This is because the equilibrium constant and reaction quotient are typically expressed in terms of atmospheres (atm) for consistency and comparison. When given pressures in units like torr, converting them to atm helps in correctly computing the reaction quotient, which directly influences the assessment of the system's equilibrium status.

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Most popular questions from this chapter

At a particular temperature, \(K=4.0 \times 10^{-7}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$ In an experiment, \(1.0 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}\) is placed in a 10.0-L vessel. Calculate the concentrations of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) when this reaction reaches equilibrium.

Consider the following statements: "Consider the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\), for which at equilibrium \([\mathrm{A}]=2 M\) \([\mathrm{B}]=1 M\), and \([\mathrm{C}]=4 M .\) To a 1 -L container of the system at equilibrium, you add 3 moles of \(B\). A possible equilibrium condition is \([\mathrm{A}]=1 M,[\mathrm{~B}]=3 M\), and \([\mathrm{C}]=6 M\) because in both cases \(K=2 .\) " Indicate everything that is correct in these statements and everything that is incorrect. Correct the incorrect statements, and explain.

As we have attempted to lessen our dependence on fossil fuels, the demand for biofuels, such as ethanol, which is produced by the fermentation of the sugars found in corn, has increased. Using Le Châtelier's principle, predict which way the equilibrium will shift during the fermentation of sugar for each of the following changes. $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(a q) $$ a. when the concentration of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is increased b. when the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) is decreased c. when \(\mathrm{CO}_{2}\) gas is added to the solution d. when the volume of water in the solution is doubled

Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\), $$ \begin{array}{ll} \mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\ \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\ \mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\ \mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14} \end{array} $$ determine the values for the equilibrium constants for the following reactions. a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C}\), equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

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