Question

The \(\mathrm{N}_{2} \mathrm{O}_{4}-\mathrm{NO}_{2}\) reversible reaction is found to have the following equilibrium concentrations at \(100^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{eq}}\) for the reaction. $$ \begin{array}{c} \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \\ 4.5 \times 10^{-5} \mathrm{M} \quad 3.0 \times 10^{-3} \mathrm{M} \end{array} $$

Short answer

The equilibrium constant, \(K_{eq}\), for the reaction is \(2.0 \times 10^{-1}\).

Step-by-step solution

Write the Expression for Equilibrium Constant

The equilibrium constant expression, \(K_{eq}\), for the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\) is given by the formula: \[K_{eq} = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\]This expression arises from the balanced chemical equation, where the concentration of each species is raised to the power of its stoichiometric coefficient.

Substitute Known Concentrations into the Expression

Substitute the given equilibrium concentrations into the expression for \(K_{eq}\):- \([\mathrm{NO}_2] = 3.0 \times 10^{-3} \mathrm{M}\)- \([\mathrm{N}_2\mathrm{O}_4] = 4.5 \times 10^{-5} \mathrm{M}\)Hence,\[K_{eq} = \frac{(3.0 \times 10^{-3})^2}{4.5 \times 10^{-5}}\]

Perform the Calculation

Calculate the value of \(K_{eq}\):- Calculate the numerator: \( (3.0 \times 10^{-3})^2 = 9.0 \times 10^{-6} \)- Substitute back into the equation: \[K_{eq} = \frac{9.0 \times 10^{-6}}{4.5 \times 10^{-5}}\]- Simplify: \[K_{eq} = \frac{9.0}{4.5} \times 10^{-6 + 5} = 2.0 \times 10^{-1}\]

Verify the Calculation

Check through the calculations and ensure each step follows logically. Re-evaluate multiplication and division to confirm:\[K_{eq} = 2.0 \times 10^{-1}\]This verifies our calculations are correct.

Key concepts

Equilibrium Constant

The equilibrium constant, represented as \(K_{eq}\), is crucial in understanding chemical equilibrium. It indicates the ratio of the concentration of products to reactants when a reaction is at equilibrium.
For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \), the equilibrium constant expression is written as:

• \[K_{eq} = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\]

To write this expression, you raise the concentration of each component to the power of its coefficient in the balanced equation.
In this particular reaction, \([\mathrm{NO}_2]\) is squared because the balanced equation indicates 2 moles of \(\mathrm{NO}_2\) per mole of reaction. Understanding \(K_{eq}\) helps predict how the reaction behaves and shifts when conditions change, such as temperature or pressure.

Stoichiometry

Stoichiometry is the heart of understanding chemical reactions, including the concept of chemical equilibrium. It involves quantifying relationships between reactants and products in a chemical reaction based on balanced equations.
In a chemical equation like \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\), stoichiometry tells us:

• One molecule of \(\mathrm{N}_{2} \mathrm{O}_{4}\) produces two molecules of \(\mathrm{NO}_2\).

The stoichiometric coefficients (like the '2' before \(\mathrm{NO}_2\)) guide how we set up the equilibrium expression \(K_{eq}\). They're also crucial when applying the law of conservation of mass, ensuring that the number of atoms remains the same before and after a reaction.
In practice, understanding stoichiometry allows you to calculate concentrations of reactants and products and, ultimately, to solve for equilibrium constants more accurately.

Reversible Reactions

Reversible reactions are those that can go in both the forward and backward direction. This is a key feature in achieving equilibrium in chemical reactions. The double arrow, \(\rightleftarrows\), in the reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \), highlights that the reaction can proceed towards products, \(\mathrm{NO}_2\), and also backward towards reactants, \(\mathrm{N}_{2} \mathrm{O}_{4}\).
In a reversible reaction, the system reaches a state where the rate of the forward reaction equals the rate of the backward reaction, meaning the concentrations of reactants and products remain constant over time.
This equilibrium state does not imply that the quantities are equal, but rather that their ratio is constant, as dictated by \(K_{eq}\). Understanding reversible reactions is crucial when predicting how different interventions, such as pressure changes or catalysts, will affect the system's equilibrium position.