Question

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{HI}(g)\), equilibrium concentrations at \(298 \mathrm{~K}\) are \(\left[\mathrm{H}_{2}\right]=0.0510 \mathrm{~mol} / \mathrm{L},\left[\mathrm{I}_{2}\right]=0.174 \mathrm{~mol} / \mathrm{L},\) and \([\mathrm{HI}]=0.507 \mathrm{~mol} / \mathrm{L} .\) What is the value of \(K\) at \(298 \mathrm{~K} ?\)

Short answer

The equilibrium constant \(K\) at 298 K is 29.0.

Step-by-step solution

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2\mathrm{HI}(g)\) is given by:\[K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2][\mathrm{I}_2]}\]where \([\mathrm{HI}]\), \([\mathrm{H}_2]\), and \([\mathrm{I}_2]\) are the equilibrium concentrations of the respective species.

Substitute the Known Values

Substitute the given equilibrium concentrations into the expression:- \([\mathrm{H}_2] = 0.0510 \; \mathrm{mol/L}\)- \([\mathrm{I}_2] = 0.174 \; \mathrm{mol/L}\)- \([\mathrm{HI}] = 0.507 \; \mathrm{mol/L}\)\[K = \frac{(0.507)^2}{(0.0510)(0.174)}\]

Calculate the Value of the Equilibrium Constant

Perform the calculation by first finding the square of \([\mathrm{HI}]\):\[(0.507)^2 = 0.257049\]Then multiply \([\mathrm{H}_2]\) and \([\mathrm{I}_2]\):\[(0.0510)(0.174) = 0.008874\]Finally, divide the squared concentration by the product of these concentrations:\[K = \frac{0.257049}{0.008874} = 28.958\]

Final Step: Round the Equilibrium Constant

Round the calculated value of the equilibrium constant to three significant figures, considering the precision of the given data, thus:\[K = 29.0\]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry, where the rate of the forward reaction equals the rate of the reverse reaction. This balance leads to a situation where the concentrations of reactants and products remain constant over time, even though the reactions are ongoing. Imagine a busy highway: cars (reactants) entering and leaving a city (products) in perfect balance. That's chemical equilibrium in a nutshell!
To describe equilibrium quantitatively, chemists use the equilibrium constant, denoted as \( K \). It is derived from the concentrations of the products divided by those of the reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation. A large value of \( K \) indicates a greater concentration of products at equilibrium, pointing to the reaction favoring the formation of products.

Concentration Calculations

In chemistry, concentration calculations are crucial for determining how much of a substance is present in a given volume. When dealing with chemical equilibrium, such calculations help us understand the system's state at given conditions. It involves using the equilibrium concentrations of all species involved.
For the reaction \( \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftarrows 2\mathrm{HI}(g) \), we're given the concentrations at equilibrium: \( [\mathrm{H}_{2}] = 0.0510 \ \mathrm{mol/L} \), \( [\mathrm{I}_{2}] = 0.174 \ \mathrm{mol/L} \), and \( [\mathrm{HI}] = 0.507 \ \mathrm{mol/L} \). These values tell us how much of each species is present at equilibrium, and they are pivotal for finding the equilibrium constant \( K \).
To ensure accuracy, pay attention to significant figures which reflect the precision of measurements. They indicate how precise your data is – crucial when calculating values like the equilibrium constant.

Reaction Quotients

Reaction quotients, denoted as \( Q \), are similar to equilibrium constants (\( K \)), but they don't require the system to be at equilibrium. \( Q \) can be calculated at any point in time during a reaction using the same formula as \( K \): the concentrations of products over reactants, each raised to their respective powers.
Here's the trick: by comparing \( Q \) and \( K \), we can predict the direction the reaction will proceed to reach equilibrium.

• If \( Q < K \), the reaction will proceed forward, forming more products.
• If \( Q > K \), the reaction will shift in reverse, creating more reactants.
• If \( Q = K \), the reaction is already at equilibrium, and the concentrations will remain constant.

Thus, \( Q \) is a powerful tool for chemists to understand reaction dynamics and adjust conditions to achieve a desired equilibrium state. It helps in predicting how changes in concentration, pressure, or temperature can affect the position of equilibrium.