Question

What will be the equilibrium constant at \(717 \mathrm{~K}\) for the reaction $$ 2 \mathrm{HI}(\mathrm{g}) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{~g}) $$ if its value for the reaction $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) \text { at } 717 \mathrm{~K} \text { is } 64 ? $$ (a) 8 (b) 64 (c) \(\frac{1}{64}\) (d) \(\frac{1}{8}\)

Short answer

\(\frac{1}{64}\)

Step-by-step solution

Identify the relationship between the reactions

The first reaction given is the reverse of the second reaction. Thus, the equilibrium constant of the first reaction would be the reciprocal of the equilibrium constant for the second reaction.

Use the equilibrium constant for the second reaction

According to the problem, the equilibrium constant for the reaction \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)\) at \(717 K\) is given to be \(64\). We denote this as \(K_2 = 64\).

Solve for the equilibrium constant of the first reaction

As explained earlier, the equilibrium constant of the reverse reaction is the reciprocal of that of the original reaction. Hence, the equilibrium constant for the first reaction (denoted as \(K_1\)) is \(\frac{1}{K_2} = \frac{1}{64}\).

Key concepts

Reversible Reactions

In chemistry, a reversible reaction is a type of chemical reaction that can proceed in both forward and backward directions. This means that the reactants can form products, and the products can also revert back to form the original reactants. Imagine it like a seesaw where the system can balance out at any given point depending on the conditions, such as temperature and pressure. Reversible reactions are typically represented with a double-headed arrow \[ \rightleftharpoons \] to signify that they can proceed in both directions. For example, the reaction between hydrogen gas and iodine gas to form hydrogen iodide can go forward or backward, making it a reversible reaction.

• The forward reaction converts hydrogen and iodine into hydrogen iodide.
• The backward reaction converts hydrogen iodide back into hydrogen and iodine.

In real-life scenarios, most chemical reactions are indeed reversible to some extent, and the concept of reversibility is crucial for understanding equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs in a reversible reaction when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time. It is like a dance where both partners move in perfect harmony, neither getting ahead nor falling behind.Though the reactions continue to occur, because they happen at the same rate, there is no net change in the amount of reactants and products. The system can be described by an equilibrium constant, denoted as \(K\). This constant is a ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

• If \(K\) is large (greater than 1), products are favored at equilibrium.
• If \(K\) is small (less than 1), reactants are favored.

The given exercise illustrates this concept: the equilibrium constant for a reaction and its reverse are reciprocally related, showcasing the delicate balance typical of chemical equilibrium.

Temperature Dependence of Equilibrium Constants

The equilibrium constant for a reaction is not a static number; it can change with temperature. This change is because temperature affects the rates of both the forward and backward reactions differently. By altering the balance between these rates, temperature shifts the equilibrium position. Understanding this concept is crucial when predicting how a reaction will behave under different thermal conditions. Le Chatelier's principle offers an easy way to predict the direction of this shift. For endothermic reactions, increasing the temperature generally leads to an increase in the equilibrium constant. For exothermic reactions, the opposite occurs—higher temperatures usually result in a decrease in the equilibrium constant. In the exercise, the equilibrium constant is given at a specific temperature of 717 K, showing that it should not be assumed valid at other temperatures without further information. This emphasizes the dynamic nature of equilibrium constants and the need for precise thermal control in chemical experimentation.