Problem 61 - 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 \(K_{\mathrm{c}}=1.7 \times 10^{-3}\). Suppose that \(0.015 \mathrm{~mol} \mathrm{NO}(\mathrm{g}), 0.25 \mathrm{~mol} \mathrm{~N}_{2}(\mathrm{~g}),\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}(\mathrm{~g})\) are placed into a \(10.0-\mathrm{L}\) flask, sealed, and heated to \(2300 \mathrm{~K}\). (a) Determine whether the system is at equilibrium. (b) If not, in which direction must the reaction proceed to reach equilibrium? (c) Calculate the equilibrium concentrations of all three substances.

Step-by-Step Solution

Answer

(a) Not at equilibrium; (b) shifts left; (c) [NO] = 0.0023 M, [N2] = 0.0246 M, [O2] = 0.0246 M.

Step 1: Calculate Initial Concentrations

First, find the initial concentrations of each gas by dividing their moles by the volume of the flask. \[ [NO]_0 = \frac{0.015 \text{ mol}}{10.0 \text{ L}} = 0.0015 \text{ M} \]\[ [N_2]_0 = \frac{0.25 \text{ mol}}{10.0 \text{ L}} = 0.025 \text{ M} \]\[ [O_2]_0 = \frac{0.25 \text{ mol}}{10.0 \text{ L}} = 0.025 \text{ M} \]

Step 2: Write the Reaction Quotient Expression

The reaction quotient \( Q_c \) is calculated from:\[ Q_c = \frac{[NO]^2}{[N_2][O_2]} \]Substitute the initial concentrations:\[ Q_c = \frac{(0.0015)^2}{0.025 \times 0.025} \]

Step 3: Calculate the Reaction Quotient

Evaluate \( Q_c \):\[ Q_c = \frac{0.00000225}{0.000625} = 0.0036 \]

Step 4: Compare \( Q_c \) to \( K_c \)

Compare \( Q_c = 0.0036 \) with \( K_c = 1.7 \times 10^{-3} \).Since \( Q_c > K_c \), the reaction is not at equilibrium and will shift to the left.

Step 5: Set Up the ICE Table

Use an ICE table to track changes in concentration:Initial: \( [NO]_0 = 0.0015 \), \( [N_2]_0 = 0.025 \), \( [O_2]_0 = 0.025 \)Change: \( [NO] = -2x \), \( [N_2] = +x \), \( [O_2] = +x \)Equilibrium: \( [NO] = 0.0015 - 2x \), \( [N_2] = 0.025 + x \), \( [O_2] = 0.025 + x \)

Step 6: Write Equilibrium Expressions

Equilibrium expression is:\[ K_c = \frac{(0.0015 - 2x)^2}{(0.025 + x)(0.025 + x)} \]Set this equal to \( 1.7 \times 10^{-3} \).

Step 7: Solve for \( x \) Using Quadratic Equation

Simplify and solve the quadratic equation:\[ (0.0015 - 2x)^2 = (0.025 + x)^2 \cdot 1.7 \times 10^{-3} \]Upon solving, we find \( x \approx -0.0004 \), but only viable within concentration constraints.

Step 8: Calculate Equilibrium Concentrations

Substitute \( x \) back to find equilibrium concentrations:\[ [NO] = 0.0015 - 2(-0.0004) = 0.0023 \text{ M} \]\[ [N_2] = 0.025 + (-0.0004) = 0.0246 \text{ M} \]\[ [O_2] = 0.025 + (-0.0004) = 0.0246 \text{ M} \]

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \), is a vital concept in chemical equilibrium that helps us understand the ratio of concentrations of products to reactants at equilibrium. This constant is temperature-dependent, meaning it remains unchanged unless the temperature changes.
The equation for an equilibrium constant for a general reaction like \( aA + bB \rightleftharpoons cC + dD \) can be given by:

• \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

At equilibrium, the reaction quotient \( Q_c \) changes its label to \( K_c \).
In the exercise, the equilibrium constant \( K_c \) value was provided as \( 1.7 \times 10^{-3} \) at a specific temperature of 2300 K. The \( K_c \) is derived from the concentration of chemical species in the reaction.
When calculating \( K_c \), only the concentrations of gaseous and aqueous species are considered because only they contribute to the equilibrium position.
Solid and liquid concentrations are constants and thus omitted from the equilibrium expression.

Reaction Quotient

The reaction quotient, denoted as \( Q_c \), serves as a snapshot of a system that can help determine whether the chemical reaction proceeds towards reactants or products.
Like \( K_c \), the formula for \( Q_c \) is the same:

• \( Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

The difference is that \( Q_c \) uses the initial concentrations of the reactants and products, while \( K_c \) uses concentrations once the system reaches equilibrium.
In the given exercise, the initial concentrations were used to calculate \( Q_c \), which turned out to be 0.0036.
Since \( Q_c > K_c \), it signals that the system had too many products at the start compared to the equilibrium point and thus needed to shift towards the reactants to reach equilibrium. This comparison is crucial as it tells you in which direction the reaction must proceed.

ICE Table

An ICE table is a helpful tool for organizing what happens in a chemical reaction in terms of concentrations. ICE stands for "Initial, Change, Equilibrium" and helps in solving equilibrium problems.
Here is how an ICE table can be structured:

• **Initial**: The starting concentrations of all species before the reaction has shifted towards equilibrium.
• **Change**: The change in concentrations as the system moves towards equilibrium. Often represented with variables like \( x \), where changes can be either positive or negative depending on direction (reactants or products are formed).
• **Equilibrium**: The concentrations of each species when equilibrium is established. These values are the initial concentrations plus or minus the changes.

For the problem, starting concentrations were found for \([NO]_0\), \([N_2]_0\), and \([O_2]_0\), using provided values. The changes were calculated using the variable \( x \), representing the amount shifted.
By applying the equilibrium condition \( K_c = 1.7 \times 10^{-3} \) and setting up the equilibrium expressions, the value of \( x \) was determined. This allowed for calculating the final equilibrium concentrations of the substances in the reaction.