Problem 53 - Chemistry The Molecular Science

Question

Consider the equilibrium $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ At \(2300 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}=1.7 \times 10^{-3}\). If \(0.15 \mathrm{~mol} \mathrm{NO}(\mathrm{g})\) is placed into an empty, sealed \(10.0-\mathrm{L}\) flask and heated to \(2300 \mathrm{~K},\) calculate the equilibrium concentrations of all three substances at this temperature.

Step-by-Step Solution

Answer

Equilibrium concentrations are: [NO] = 0.0147 M, [N2] = 9.8e-4 M, [O2] = 9.8e-4 M.

Step 1: Write the Expression for the Equilibrium Constant

For the given reaction, the expression for the equilibrium constant, \( K_c \), is: \[ K_c = \frac{[\text{NO}]^2}{[\text{N}_2][\text{O}_2]} \]where \([\text{NO}], [\text{N}_2], [\text{O}_2]\) are the concentrations of NO, N\(_2\), and O\(_2\) at equilibrium, respectively.

Step 2: Determine Initial Concentrations

Initially, only NO is present with a concentration of\[ [\text{NO}]_0 = \frac{0.15 \, \text{mol}}{10.0 \, \text{L}} = 0.015 \, \text{M} \]N\(_2\) and O\(_2\) have initial concentrations of 0 M.

Step 3: Define Changes in Concentrations

Assume that \( x \) moles of N\(_2\) and O\(_2\) form at equilibrium. Then, 2\( x \) \, moles of NO will dissociate. Therefore, the equilibrium concentrations will be:\[ [\text{NO}] = 0.015 - 2x \]\[ [\text{N}_2] = x \]\[ [\text{O}_2] = x \]

Step 4: Substitute and Solve for \( x \)

Substitute the equilibrium concentrations back into the \( K_c \) expression:\[ 1.7 \times 10^{-3} = \frac{(0.015 - 2x)^2}{x^2} \]This simplifies to solve for \( x \). It yields a quadratic equation:\[ (0.015 - 2x)^2 = 1.7 \times 10^{-3} x^2 \]

Step 5: Simplify and Rearrange the Equation

Expanding the left side:\[ 0.000225 - 0.06x + 4x^2 = 1.7 \times 10^{-3} x^2 \]Rearrange to form a standard quadratic equation:\[ 2.3x^2 - 0.06x + 0.000225 = 0 \]

Step 6: Solve the Quadratic Equation

Use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \( a = 2.3, \, b = -0.06, \, c = 0.000225 \).Calculate to find the values of \( x \).

Step 7: Calculate Equilibrium Concentrations

Substitute the value of \( x \) back into the expressions for equilibrium concentrations:\[ [\text{NO}] = 0.015 - 2x \]\[ [\text{N}_2] = x \]\[ [\text{O}_2] = x \]Calculate the final concentrations using the value of \( x \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a central concept in the study of chemical reactions. It occurs when the forward and reverse reaction rates are equal, leading to constant concentrations of reactants and products over time. Although reactions are dynamic at the molecular level, with ongoing exchanges between reactants and products, the concentrations of each substance remain stable in equilibrium. This stability is achieved because the two opposing processes cancel each other out.

In the context of the equilibrium expression given by the reaction \[\mathrm{N}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \, \mathrm{NO}(\mathrm{g})\]the equilibrium constant, \(K_c\), plays a crucial role. This constant allows us to understand the relative amounts of reactants and products at equilibrium. A high \(K_c\) value indicates that, at equilibrium, products dominate over reactants. Conversely, a low \(K_c\) suggests reactants are favored.

Here, with \(K_{c} = 1.7 \times 10^{-3}\), it indicates a reaction heavily tilted towards the left, meaning that at equilibrium, you will find lesser quantities of NO compared to unreacted \(\mathrm{N}_2\) and \(\mathrm{O}_2\).

Concentration

Concentration refers to the amount of a substance in a given volume of solution and is typically expressed in molarity (mol/L). For chemical reactions, knowing the initial concentrations of reactants and products is crucial in determining the position of equilibrium.

In the given problem, we start with the concentration of \(\mathrm{NO}\) being \[[\text{NO}]_0 = \frac{0.15 \, \text{mol}}{10.0 \, \text{L}} = 0.015 \, \text{M}\]at the beginning of the reaction. The concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both initially zero since they are not present initially.

Keeping track of concentration changes during a reaction is vital. As the system reaches equilibrium, the concentration of \(\mathrm{NO}\) will decrease due to the reaction shifting to the left (forming more \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\)). Meanwhile, the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) will increase from zero to a positive value, \(x\), which can be calculated at equilibrium.

Quadratic Equation

In chemical equilibrium problems, solving requires manipulating equations to find unknowns, which often leads us to quadratic equations. Quadratic equations, in their most general form, are expressed as \[ax^2 + bx + c = 0\]where \(a\), \(b\), and \(c\) are constants.

This specific problem leads to a quadratic equation of the form:\[2.3x^2 - 0.06x + 0.000225 = 0\] This equation arises after substituting the equilibrium concentrations back into the expression for \(K_c\).

To solve such a quadratic equation, the quadratic formula is used:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In this case, \(a = 2.3\), \(b = -0.06\), and \(c = 0.000225\). Solving this allows us to determine the value of \(x\), which represents the concentration changes of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) at equilibrium. By plugging the \(x\) values back, we find the equilibrium concentrations of all species involved.