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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 \(\left[I_{2}\right]=4.79 \times 10^{-4} M\). What is the value of \(K_{c}\) at this temperature?

Short Answer

Expert verified
The value of the equilibrium constant, \(K_c\), for the given reaction at this temperature is 18.46.

Step by step solution

01

1. Write the balanced chemical equation and the Kc expression.

The balanced chemical equation is given by: \(2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)\) The equilibrium constant (Kc) expression for this reaction is: \[K_c = \frac{[H_2][I_2]}{[HI]^2}\]
02

2. Plug in the equilibrium concentrations.

We are given the following equilibrium concentrations: \[ [HI] = 3.53 \times 10^{-3} M \] \[ [H_2] = 4.79 \times 10^{-4} M \] \[ [I_2] = 4.79 \times 10^{-4} M \] Now, plug these concentrations into the Kc expression: \[K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2}\]
03

3. Calculate the value of Kc.

To find the value of Kc, simply perform the calculations: \[K_c = \frac{(4.79 \times 10^{-4})(4.79 \times 10^{-4})}{(3.53 \times 10^{-3})^2} = 18.46\] The value of Kc at this temperature is 18.46.

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Key Concepts

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

Equilibrium Constant Expression
In the realm of chemical reactions, the equilibrium constant expression serves as a mathematical representation of a chemical reaction's equilibrium state. When we refer to a chemical reaction reaching equilibrium, we mean that the rates of the forward and reverse reactions are equal, leading to constant concentrations of the reactants and products over time.

This expression is denoted by the symbol, \(K_c\) for reactions in solution or \(K_p\) if we are dealing with gases and partial pressures. The equilibrium constant is a crucial tool, as it provides insight into the extent of a reaction; the higher the value of \(K_c\), the more products are formed, which implies that the reaction favors the formation of products over reactants.

The general form of the equilibrium constant expression for a reaction like \(aA + bB \rightleftharpoons cC + dD\) is given by the formula: \[K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] where the square brackets represent the concentrations of the substances (in moles per liter), and the letters \(a\), \(b\), \(c\), and \(d\) correspond to the stoichiometric coefficients from the balanced reaction equation.

In our example with the decomposition of hydrogen iodide, the reaction is: \(2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)\), giving us the equilibrium constant expression: \[K_c = \frac{[H_2][I_2]}{[HI]^2}\] This expression allows us to quantify the position of equilibrium based on the measured concentrations of the reactants and products at a given temperature.
Equilibrium Concentrations
Equilibrium concentrations are the amounts of reactants and products present in a reaction mixture when the reaction has reached equilibrium. At this point, no observable changes occur in the system's composition, as the rate of the forward reaction equals that of the reverse reaction. Understanding and calculating these concentrations is essential in predicting how a reaction mixture will respond to changes in conditions, like temperature or pressure, according to Le Chatelier's principle.

To calculate the value of the equilibrium constant \(K_c\), we must know the equilibrium concentrations of the reactants and products. These are typically expressed in molarity (\(M\)), which is moles of substance per liter of solution. In the textbook problem presented, the concentrations are: \[ [HI] = 3.53 \times 10^{-3} M \] \[ [H_2] = 4.79 \times 10^{-4} M \] \[ [I_2] = 4.79 \times 10^{-4} M \] By using these equilibrium concentrations, we can then calculate the equilibrium constant for the system, which tells us the extent to which the reactants convert into products under the given conditions.
Chemical Reaction Equilibrium
Chemical reaction equilibrium is a condition wherein the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It's a dynamic state; even though there is no change in the overall amounts of substances, the reactants continue to form products, and the products continue to revert to reactants.

A chemical system's tendency to reach equilibrium is universal; however, the position of equilibrium, represented by the equilibrium constant, is specific to each reaction and depends on temperature. A high equilibrium constant indicates a high concentration of products at equilibrium, suggesting a reaction that largely goes to completion. Conversely, a low equilibrium constant suggests the equilibrium mixture contains relatively more reactants.

In the case of the decomposition of hydrogen iodide, the reaction reaches equilibrium when the rates at which hydrogen iodide decomposes into hydrogen and iodine and the rate at which hydrogen and iodine recombine to form hydrogen iodide become equal. The equilibrium constant calculated from the equilibrium concentrations, in this case, provides valuable information about the system's behavior under the conditions of the experiment.

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Most popular questions from this chapter

Consider \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) \(\Delta H=-904.4 \mathrm{~kJ}\). How does each of the following changes affect the yield of \(\mathrm{NO}\) at equilibriunt? Answer increase, decrease, or no change: (a) increase [NII \(_{3}\) ]; (b) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;\) (c) decrease \(\left[\mathrm{O}_{2}\right.\) \\} (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO}\), and \(\mathrm{CO}_{2}\) is placed in a 10.0-L vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO}\), and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO}\), and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (c) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO}\), and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\).

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at \(500 \mathrm{~K}\). An equilibrium mixture of the three gases has partial pressures of \(0.095 \mathrm{~atm}, 0.171 \mathrm{~atm}\), and \(0.28\) atm for \(\mathrm{NO}, \mathrm{Cl}_{2}\), and \(\mathrm{NOCl}\), respectively. Calculate \(K_{p}\) for this reaction at \(500 \mathrm{~K}\).

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\)

(a) At \(1285{ }^{\circ} \mathrm{C}\) the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3} .\) A \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel? (b)For the reaction \(\mathrm{H}_{2}(g)+\mathrm{l}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g), K_{c}=55.3\) at \(700 \mathrm{~K}\). In a 2.00-L flask containing an equilibrium mixture of the three gases, there are \(0.056 \mathrm{~g} \mathrm{H}_{2}\) and \(4.36 \mathrm{~g} \mathrm{I}_{2}\). What is the mass of HI in the flask?

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