For the reaction \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\), \(K_{\mathrm{c}}=160\) at \(500 \mathrm{~K}\). An analysis of a reaction mixture at \(500 \mathrm{~K}\) showed that it had the composition \(4.8 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}, 2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{I}_{2}\), and \(2.4 \times 10^{-3} \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{HI}\). (a) Calculate the reaction quoticnt. (b) Is the reaction mixture at equilibrium? (c) If not, is there a tendency to form more reactants or more products?

Short Answer

Expert verified
Q = 0.5, which is less than Kc (160). Therefore, the reaction mixture is not at equilibrium and there's a tendency to form more products.

Step by step solution

01

Understanding the Reaction Quotient (Q)

The reaction quotient Q is a measure of the relative amounts of products and reactants present during a reaction at a given instant. It is calculated using the same form as the equilibrium constant, Kc, but with the concentrations (or partial pressures) of the reactants and products at the time of interest rather than at equilibrium. For the given reaction, Q is computed by the formula: Q = [HI]^2 / ([H2][I2]).
02

Calculating the Reaction Quotient (Q)

Plug in the concentrations of the reactants and products into the formula for Q: Q = (2.4 x 10^-3)^2 / (4.8 x 10^-3 * 2.4 x 10^-3) = 5.76 x 10^-6 / (1.152 x 10^-5). Simplify the expression to find the value of Q.
03

Comparing Q to Kc

With Q calculated, compare its value to the equilibrium constant, Kc. If Q = Kc, the system is at equilibrium. If Q < Kc, the reaction will proceed forward, forming more products. If Q > Kc, the reaction will proceed in reverse, forming more reactants.
04

Determining the Tendency of the Reaction

By having both the value for Q and the given equilibrium constant Kc, determine the direction in which the reaction has a tendency to proceed in order 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
Chemical equilibrium is central to understanding how chemical reactions occur in nature and industry. It represents a state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of reactants and products remain constant over time.

In a closed system, once equilibrium is reached, no net change in the concentration of reactants and products can be observed, even though the reactions continue to occur at a molecular level. It's a dynamic state, not static, as molecules are constantly reacting, just at equal rates in both directions.

The concept of equilibrium does not imply that the reactants and products are present in equal amounts; rather, their ratios are fixed and depend on the reaction specifics and conditions, such as temperature and pressure. This state of balance can be disturbed by changes in these conditions, leading to what we call a shift in equilibrium. Here, we introduce the idea of reaction quotient (Q) as a way to determine how the system will adjust to regain equilibrium.
Equilibrium Constant
The equilibrium constant, denoted as K or Kc when referring to concentrations, is a number that quantifies the ratio of product to reactant concentrations at chemical equilibrium for a particular reaction at a constant temperature.

The magnitude of Kc indicates whether the reactants or products are favored in the equilibrium state. A large equilibrium constant (much greater than 1) suggests that the products are greatly favored over the reactants, meaning the forward reaction is almost complete. Conversely, a small value (much less than 1) means that the reactants are favored, and the reaction does not proceed far before reaching equilibrium.

In the provided exercise, you calculated the reaction quotient Q, which, when compared to the equilibrium constant, will reveal whether the system is at equilibrium or which direction it needs to proceed to reach that state. It's important to remember that Kc is constant for a given reaction at a specific temperature, illustrating the principle that equilibrium properties are determined by the nature of the reaction and its conditions, not by the amounts of reactants and products initially combined.
Le Châtelier's Principle
Le Châtelier's principle is an invaluable tool in predicting how a chemical system at equilibrium will respond to external changes. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the position of equilibrium will shift to counteract the change.

For example, if the concentration of a reactant is increased, the equilibrium will shift towards the products to reduce the added reactant. Similarly, a decrease in the concentration of a product will shift equilibrium towards forming more of that product. Temperature changes also impact equilibrium: increasing the temperature favors the endothermic reaction direction, while decreasing temperature favors the exothermic direction.

Le Châtelier's principle can be used to optimize chemical processes by adjusting conditions to favor the production of desired products. In industrial settings, mastering this principle allows for increased yields and efficiency. Understanding how the system at hand will respond to such adjustments is crucial for anyone involved in chemical reactions on both a theoretical and practical level.

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Most popular questions from this chapter

The value of the equilibrium constant \(K\) for the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) is \(3.0 \times 10^{4}\) at \(700 \mathrm{~K}\). Determine the value of \(K_{c}\) for the reactions (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) and \((\mathrm{b}) \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\)

\( \mathrm{Ar} 500^{\circ} \mathrm{C}, K_{\mathrm{c}}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g}) .\) Calculate the value of \(K_{\varepsilon}\) at \(500^{\circ} \mathrm{C}\) for the following reactions. (a) \(\frac{1}{6} \mathrm{~N}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \frac{1}{3} \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{NH}_{3}(g)=\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{H}_{2}(g)\) (c) \(4 \mathrm{~N}_{2}(\mathrm{~g})+12 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 8 \mathrm{NH}_{3}(\mathrm{~g})\)

Explain what is wrong with the following statements. (a) Once a reaction has reached equilibrium, all reaction stops. (b) If more reactant is used, the equilibrium constant will have a larger value.

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

A mixture consisting of \(1.1 \mathrm{mmol} \mathrm{SO}_{2}\) and \(2.2 \mathrm{mmol} \mathrm{O}_{2}\) in a \(250-\mathrm{mL}\) container was heated to \(500 \mathrm{~K}\) and allowed to reach equilibrium. Will more sulfur trioxide be formed if that equilibrium mixture is cooled to \(25^{\circ} \mathrm{C}\) ? For the reaction \(2 \mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), \mathrm{K}=2.5 \times 10^{10}\) at \(500 \mathrm{~K}\) and \(4.0 \times 10^{24}\) at \(25^{\circ} \mathrm{C}\).

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