Gaseous hydrogen iodide is placed in a closed container at \(425^{\circ} \mathrm{C}\) , where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) .\) At equilibrium it is found that \([\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=4.79 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=4.79 \times 10^{-4} \mathrm{M} .\) What is the value of \(K_{c}\) at this temperature?

Short Answer

Expert verified
The value of the equilibrium constant (Kc) for the reaction \( 2HI(g) \rightleftharpoons H_2(g)+I_2(g) \) at \(425^{\circ}C\) is approximately 0.0185, using the given equilibrium concentrations of the species (HI, H₂, and I₂).

Step by step solution

01

Write the equilibrium constant expression.

The first step is to write down the equilibrium constant formula for the given reaction. For the reaction \( 2HI(g) \rightleftharpoons H_2(g)+I_2(g) \), the equilibrium constant expression (Kc) can be written as: \[ K_c = \frac{[H_2][I_2]}{[HI]^2} \] where [H₂], [I₂], and [HI] represent the molar equilibrium concentrations of hydrogen, iodine, and hydrogen iodide, respectively.
02

Plug in the given concentrations for each substance.

We are given the following equilibrium concentrations: \[ [\mathrm{HI}]=3.53 \times 10^{-3} \mathrm{M} \\ [\mathrm{H}_{2}]=4.79 \times 10^{-4} \mathrm{M} \\ [\mathrm{I}_{2}]=4.79 \times 10^{-4} \mathrm{M} \] Now we can plug these values into our equilibrium constant expression: \[K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \]
03

Calculate Kc.

Now we can simply calculate Kc using the provided concentrations: \[K_c = \frac{(4.79 \times 10^{-4} \times 4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} \approx 0.0185 \] So the value of the equilibrium constant, Kc, at this temperature is approximately 0.0185.

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction occurs in a closed system and the reactants and products are no longer changing in concentration, the system has reached a state known as chemical equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, meaning that the reactants and products are being formed at the same rate they are being consumed.

This equilibrium does not imply that the reactants and products are present in equal amounts; rather, it indicates that their ratios are constant over time. Knowing this, students can better understand that when a reaction is at equilibrium, it is a dynamic process - one where the reactants and products are in balance, not at rest.

In the classroom or during self-study, visualizing this process with graphs or animations where the rates of the forward and backward reactions are shown to be the same can be quite helpful in grasping this concept. Furthermore, interactive simulations showing the adjustment of a reaction when disturbed can enhance the comprehension of equilibrium dynamics, also known as Le Chatelier's Principle.
Equilibrium Concentrations
The term equilibrium concentrations refers to the concentrations of the reactants and products in a chemical reaction system at the state of equilibrium. These concentrations are not necessarily equal, as they are determined by the nature of the reaction and the conditions under which it is conducted, such as temperature and pressure.

Equilibrium concentrations can be experimentally determined by measuring the concentration of each species in the reaction when it has reached equilibrium. In the context of our example problem, the hydrogen iodide (HI), hydrogen (H₂), and iodine (I₂) have specific concentrations when the system reaches equilibrium.

For students to master the concept of equilibrium concentrations, practicing with various given values and calculating the equilibrium constants (Kc) can offer deeper insight into how these concentrations influence the state of a chemical system. In fact, knowing the equilibrium concentrations allows chemists to predict the direction of the reaction under different conditions.
Reaction Quotient
The reaction quotient, denoted as Q, is a measure of the relative amounts of reactants and products at any point during a reaction before the system has reached equilibrium. Comparing the reaction quotient (Q) to the equilibrium constant (K) tells us the direction in which the reaction will proceed to reach equilibrium.

If Q < K, the reaction will proceed forward, converting reactants into products. Conversely, if Q > K, the reaction will proceed in the reverse direction, producing more reactants from products. When Q = K, the reaction is at equilibrium and will proceed at neither direction.

To calculate Q, one uses the same expression as K but with the current concentrations instead of the equilibrium concentrations. Practicing calculating Q and comparing it with K can aid students in predicting reaction behavior and becomes instrumental when dealing with systems not yet at equilibrium. Application examples from industrial processes, where maintaining a certain Q is crucial, can also provide a practical perspective on the importance of the reaction quotient.

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Most popular questions from this chapter

Which of the following statements are true and which are false? (a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow, the equilibrium constant has a value that is very close to zero. (c) As the value of the equilibrium constant increases, the speed at which a reaction reaches equilibrium increases.

Consider the following equilibrium between oxides of nitrogen $$3 \mathrm{NO}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}(g)$$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) for this reaction. (b) Will the equilibrium constant for the reaction increase or decrease with increasing temperature? (c) At constant temperature, would a change in the volume of the container affect the fraction of products in the equilibrium mixture?

How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in the volume, (d) decrease in the temperature, (e) addition of a catalyst?

At a temperature of 700 \(\mathrm{K}\) , the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at 700 \(\mathrm{K} ?\) (b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{s}^{-1}\) at 800 \(\mathrm{K} ?\)

At \(373 \mathrm{K}, K_{p}=0.416\) for the equilibrium $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressures of \(\mathrm{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g) ?\)

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