Question

The value of the equilibrium constant for the reaction below is 40.0 at a specified temperature. What would be the value of that constant for the reverse reaction under the same conditions? $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)$$

Short answer

0.025

Step-by-step solution

Identify the given equilibrium constant

The equilibrium constant for the reaction \(\text{H}_2(g) + \text{I}_2(g) \rightleftarrows 2 \text{HI}(g)\) is given as 40.0.

Understand the concept of equilibrium constants for reverse reactions

For a reaction, if the equilibrium constant of the forward reaction is \(K_f\), then the equilibrium constant of the reverse reaction \(K_r\) is the reciprocal of \(K_f\). Mathematically, this is represented as \(K_r = \frac{1}{K_f}\).

Apply the formula

Using the formula for the reverse reaction's equilibrium constant, calculate \(K_r = \frac{1}{40.0}\).

Perform the calculation

Calculate \( \frac{1}{40.0} \) to get the value of the equilibrium constant for the reverse reaction. \( K_r = 0.025 \).

Key concepts

Reversible Reactions

In chemistry, many reactions don't just go one way, they can happen in both directions. This is what we call a reversible reaction. A reversible reaction can go forward to create products from reactants or go backward to create reactants from products. For example, in our given reaction, hydrogen gas (\text{H}\(_2\)) reacts with iodine gas (\text{I}\(_2\)) to form hydrogen iodide (\text{HI}), but at the same time, hydrogen iodide can break down into hydrogen and iodine.

We use the double arrow ( \rightleftarrows ) to show that a reaction is reversible. This means the system can reach a state where both the reactants and products are present, and neither side completely consumes the other.

Reversible reactions are crucial in understanding how chemical systems behave because they allow us to predict the proportions of reactants and products under different conditions.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, though they are not necessarily equal. The point of equilibrium is dynamic—both reactions continue to occur, but they happen at the same rate, canceling each other out.

To understand chemical equilibrium, let's look at the example of the reaction: \text{H}\(_2\)(g) + \text{I}\(_2\)(g) \rightleftarrows 2 \text{HI}(g). Initially, if we start with only hydrogen and iodine gases, there will be no \text{HI}. As the reaction proceeds, \text{HI} is formed, and its concentration increases. Simultaneously, the reverse reaction starts to form \text{H}\(_2\) and \text{I}\(_2\) from \text{HI}.

Eventually, the system reaches a point where the forward reaction rate (formation of \text{HI}) is equal to the reverse reaction rate (decomposition of \text{HI}). This balance point is the chemical equilibrium, where the concentration of \text{H}\(_2\), \text{I}\(_2\), and \text{HI} stays constant.

The position of the equilibrium is described by the equilibrium constant (\text{K}). For the given forward reaction, \text{K} is given as 40.0. If the forward reaction is favored, we expect more products. Conversely, if the reverse reaction is favored, we expect more reactants.

Reaction Kinetics

Reaction kinetics is the study of the rates at which chemical reactions occur and the factors affecting these rates. It helps us understand how fast a reaction reaches equilibrium and what influences this speed.

Several factors can affect reaction rates including:

• Concentration of reactants: Higher concentrations usually increase the rate because more molecules are available to collide and react.
• Temperature: Increasing temperature typically speeds up reactions by providing more energy to the molecules.
• Catalysts: Substances that increase the reaction rate without being consumed in the process. They provide an alternative pathway with a lower activation energy.
• Surface area: For reactions involving solids, a greater surface area means more particles are available to react, increasing the rate.

In our equilibrium scenario involving \text{H}\(_2\), \text{I}\(_2\), and \text{HI}, knowing the kinetics helps predict not just the final state (equilibrium) but also how fast the system will reach that state.

If we understand the kinetics, we can control and optimize the conditions to make reactions happen faster or slower as needed. This is valuable in industrial processes where maximizing efficiency and yield is crucial.