Consider the following equilibrium process at \(700^{\circ} \mathrm{C}\) $$ 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) $$ Analysis shows that there are 2.50 moles of \(\mathrm{H}_{2}, 1.35 \times\) \(10^{-5}\) mole of \(\mathrm{S}_{2}\), and 8.70 moles of \(\mathrm{H}_{2} \mathrm{~S}\) present in a 12.0-L flask at equilibrium. Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the reaction.

Short Answer

Expert verified
The equilibrium constant \(K_{\mathrm{c}}\) for the reaction is computed by substituting the calculated concentrations into the \(K_{\mathrm{c}}\) expression and performing the arithmetic operations.

Step by step solution

01

Understanding the equation

Rewrite the given equilibrium reaction: 2 H2(g) + S2(g) <-> 2 H2S(g). The coefficients in this balanced equation will be used in the calculation of \(K_{\mathrm{c}}\).
02

Calculation of concentrations

Each equilibrium concentration (in molarity, M) is calculated by dividing the number of moles of each substance by the volume of the container. For \(\mathrm{H}_{2}\), it is \(2.50 \, \text{moles} \, / \, 12.0\, \text{liters}\), for \(\mathrm{S}_{2}\) it is \(1.35 \times 10^{-5}\, \text{moles} \, / \, 12.0\, \text{liters}\) and for \(\mathrm{H}_{2} \mathrm{~S}\) it is \(8.70 \, \text{moles} \, /\, 12.0\, \text{liters}\).
03

Apply the equilibrium constant expression

The expression for \(K_{\mathrm{c}}\) is \([H_{2}S]^{2}/([H_{2}]^{2} \times [S_{2}])\), where each concentration is raised to the power of its respective coefficient in the balanced equation.
04

Substituting values into the equilibrium constant expression

Now substitute the calculated concentrations into the \(K_{\mathrm{c}}\) expression and compute the value.
05

Result interpretation

The resultant value represents the equilibrium constant of the reaction at the given temperature and reflects how far the reaction goes before achieving equilibrium.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no overall change in the concentrations of the reactants and products over time. It's essential to note that at equilibrium, reactions have not stopped; they are dynamically occurring but at rates that cancel each other out.

This balance is represented mathematically by the equilibrium constant, which provides a ratio of the concentrations of products to reactants, raised to the power of their stoichiometric coefficients. In a classroom setting, experiments often involve adjusting conditions such as concentration, temperature, or pressure and observing the system's response to understand equilibrium better.
Reaction Quotient
The reaction quotient, Q, is a measure that determines the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, but with the initial concentrations of the reactants and products instead of their equilibrium concentrations.

When comparing Q to the equilibrium constant, K, if Q < K, the reaction will proceed forward to reach equilibrium. Conversely, if Q > K, the reaction will shift in the reverse direction to attain balance. This concept allows students to predict reaction behavior even before the system reaches equilibrium, which can be particularly illuminating during laboratory exercises.
Le Chatelier's Principle
Le Chatelier's principle is a fundamental concept in chemistry that states if a dynamic equilibrium is disturbed by changing the conditions, the system responds in a way that counteracts the disturbance and re-establishes equilibrium.

For instance, if the concentration of a reactant is increased, the system will shift to consume that excess reactant by making more product, thus leaning to the right in the equilibrium expression. Le Chatelier's principle helps us understand how changing pressures, volumes, and concentrations can affect chemical reactions and equilibrium positions, making it a crucial idea for students to grasp in various scientific contexts.
Equilibrium Constant Expression
The equilibrium constant expression for a chemical reaction is derived from the balanced chemical equation and provides the relationship between the concentrations of the reactants and products at equilibrium. For a generalized reaction aA + bB <-> cC + dD, the equilibrium constant expression takes the form \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \), where the square brackets denote the molar concentration of each species, and the letters a, b, c, and d are the stoichiometric coefficients from the balanced equation.

This expression is pivotal in any equilibrium calculation, as it allows us to quantify the position of equilibrium and the extent to which reactants are converted into products under certain conditions. It is key for students to understand that the expression is unique for each reaction and constant at a given temperature.
Molarity
Molarity is the measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution, often expressed as moles per liter (mol/L or M). It's a cornerstone concept in chemistry because many reaction rates, calculations involving equilibrium, and colligative properties depend on the concentration of the reactants and products.

Calculating molarity is a basic skill for chemistry students, involving the division of the amount of substance (in moles) by the volume of the solution (in liters). This measurement makes it possible to discuss and compare chemical concentrations in a standardized way, which is particularly helpful when interpreting equilibrium constants or reaction quotients in the context of classroom experiments or problem-solving exercises.

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Most popular questions from this chapter

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 \mathrm{M},\left[\mathrm{N}_{2}\right]=0.11 \mathrm{M}\) and \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M}\). Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)

Iodine is sparingly soluble in water but much more so in carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\). The equilibrium constant, also called the partition coefficient, for the distribution of \(\mathrm{I}_{2}\) between these two phases $$ \mathrm{I}_{2}(a q) \rightleftharpoons \mathrm{I}_{2}\left(\mathrm{CCl}_{4}\right) $$ is 83 at \(20^{\circ} \mathrm{C}\). (a) A student adds \(0.030 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\) to \(0.200 \mathrm{~L}\) of an aqueous solution containing \(0.032 \mathrm{~g}\) \(\mathrm{I}_{2} .\) The mixture is shaken and the two phases are then allowed to separate. Calculate the fraction of \(\mathrm{I}_{2}\) remaining in the aqueous phase. (b) The student now repeats the extraction of \(I_{2}\) with another 0.030 \(\mathrm{L}\) of \(\mathrm{CCl}_{4}\). Calculate the fraction of the \(\mathrm{I}_{2}\) from the original solution that remains in the aqueous phase. (c) Compare the result in (b) with a single extraction using \(0.060 \mathrm{~L}\) of \(\mathrm{CCl}_{4}\). Comment on the difference.

At \(25^{\circ} \mathrm{C}\), the equilibrium partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.15 \mathrm{~atm}\) and \(0.20 \mathrm{~atm}\), respectively. If the volume is doubled at constant temperature, calculate the partial pressures of the gases when a new equilibrium is established.

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$ \mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g) $$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C}\), respectively.)

Define homogeneous equilibrium and heterogeneous equilibrium. Give two examples of each.

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