Question

Even at high \(T,\) the formation of \(\mathrm{NO}\) is not favored: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) \quad K_{\mathrm{c}}=4.10 \times 10^{-4} \text {at } 2000^{\circ} \mathrm{C} $$ What is [NO] when a mixture of \(0.20 \mathrm{~mol}\) of \(\mathrm{N}_{2}(g)\) and \(0.15 \mathrm{~mol}\) of \(\mathrm{O}_{2}(g)\) reaches equilibrium in a \(1.0-\mathrm{L}\) container at \(2000^{\circ} \mathrm{C} ?\)

Short answer

The equilibrium concentration of \( \mathrm{NO} \) is approximately \(0.00348 \text{ M}\).

Step-by-step solution

Write the Expression for the Equilibrium Constant

The equilibrium constant expression for the reaction can be written as:\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \] where \[ K_c = 4.10 \times 10^{-4} \] and [NO], [N_2], and [O_2] are the equilibrium concentrations of the respective species.

Define Initial Concentrations

Given that we have a 1.0 L container, the initial concentrations are the same as the initial moles:\[ [N_2]_0 = 0.20 \text{ M} \] \[ [O_2]_0 = 0.15 \text{ M} \] and initially, \[ [NO]_0 = 0 \text{ M} \] since no \( \mathrm{NO} \) is present initially.

Define Changes in Concentrations

Let the change in concentration of \( \mathrm{NO} \) be \( +2x \). Given the stoichiometry, the changes in \( [N_2] \) and \( [O_2] \) will be \( -x \) each.Thus, at equilibrium, the concentrations will be:\[ [N_2] = 0.20 - x \] \[ [O_2] = 0.15 - x \] \[ [NO] = 2x \]

Set Up the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium constant expression:\[ 4.10 \times 10^{-4} = \frac{(2x)^2}{(0.20 - x)(0.15 - x)} \] Solve for \( x \).

Simplify and Solve the Quadratic Equation

First, simplify the expression:\[ 4.10 \times 10^{-4} = \frac{4x^2}{(0.20 - x)(0.15 - x)} \]\[ 4.10 \times 10^{-4} \times (0.03 - 0.35x + x^2) = 4x^2 \]\[ 4.10 \times 10^{-4} \times 0.03 - 4.10 \times 10^{-4} \times 0.35x + 4.10 \times 10^{-4} \times x^2 = 4x^2 \]\[ 1.23 \times 10^{-5} - 1.435 \times 10^{-4}x + 4.10 \times 10^{-4}x^2 = 4x^2 \] Combine like terms into a standard quadratic form \( ax^2 + bx + c = 0 \):\[ 4x^2 - 4.10 \times 10^{-4}x^2 + 1.435 \times 10^{-4}x - 1.23 \times 10^{-5} = 0 \]\[ (4 - 4.10 \times 10^{-4})x^2 + 1.435 \times 10^{-4}x - 1.23 \times 10^{-5} = 0 \]\[ 3.99959x^2 + 1.435 \times 10^{-4}x - 1.23 \times 10^{-5} = 0 \] Solve the quadratic using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \[ a = 3.99959 \] \[ b = 1.435 \times 10^{-4} \]\[ c = -1.23 \times 10^{-5} \].

Calculate and Interpret the Roots

Calculate the discriminant:\[ b^2 - 4ac = (1.435 \times 10^{-4})^2 - 4(3.99959)(-1.23 \times 10^{-5}) \]\[ = 2.059225 \times 10^{-8} + 1.967 \times 10^{-4} \]\[ \approx 1.967 \times 10^{-4} \]Thus, \[ x = \frac{-1.435 \times 10^{-4} \pm \sqrt{1.967 \times 10^{-4}}}{2(3.99959)} \]We need the positive root since \( x \) represents a concentration change, so\[ x \approx \frac{-1.435 \times 10^{-4} + 1.4025 \times 10^{-2}}{7.99918} \]\[ x \approx \frac{1.3885 \times 10^{-2}}{7.99918} \] \[ \approx 0.00174 \]

Determine the Equilibrium Concentration of NO

Finally, the equilibrium concentration of \( \mathrm{NO} \) is given by:\[ [NO] = 2x \] \[ [NO] = 2(0.00174) \]\[ [NO] \approx 0.00348 \text{ M} \]

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \( K_c \), is a crucial concept in chemical equilibrium that helps us understand the ratio of concentrations of reactants and products at equilibrium. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant expression is: \[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]This equation shows the relationship between the concentrations of the reactants \( A \) and \( B \), and the products \( C \) and \( D \) at equilibrium. For the reaction in the exercise, \( \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \), the equilibrium constant expression is:\[ K_c = \frac{[NO]^2}{[N_2][O_2]} \]Given \( K_c = 4.10 \times 10^{-4} \) at 2000°C, this tells us that at equilibrium, the product of the concentrations of \( \mathrm{NO} \) squared divided by the product of the concentrations of \( \mathrm{N_2} \) and \( \mathrm{O_2} \) equals \( 4.10 \times 10^{-4} \). Starting with 0.20 moles of \( \mathrm{N_2} \) and 0.15 moles of \( \mathrm{O_2} \) in a 1.0 L container, we aim to find the concentration of \( \mathrm{NO} \) at equilibrium.

Quadratic Equation

Solving the equilibrium problem involves using a quadratic equation because we need to determine how the concentrations of reactants and products change to reach equilibrium. The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \]From the equilibrium expression and the initial conditions, we set up a quadratic equation: \[ 4.10 \times 10^{-4} = \frac{4x^2}{(0.20 - x)(0.15 - x)} \]Simplifying, we get: \[ 4.10 \times 10^{-4} \times (0.03 - 0.35x + x^2) = 4x^2 \]We then collect like terms and form a standard quadratic equation: \[ 3.99959x^2 + 1.435 \times 10^{-4}x - 1.23 \times 10^{-5} = 0 \]To solve this, we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By substituting \( a = 3.99959 \), \( b = 1.435 \times 10^{-4} \), and \( c = -1.23 \times 10^{-5} \), we calculate the roots. This process yields the positive root representing the concentration change, \( x \), which leads us to obtain the equilibrium concentration of \( \mathrm{NO} \).

Stoichiometry

Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. In this specific problem, it allows us to determine the changes in molar concentrations of reactants and products as they reach equilibrium. For the reaction \( \mathrm{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)} \), the stoichiometric coefficients are 1:1:2.This means that for every 1 mole of \( \mathrm{N_2} \) and 1 mole of \( \mathrm{O_2} \) that react, 2 moles of \( \mathrm{NO} \) are formed. During the reaction, if \( x \) moles of \( \mathrm{N_2} \) and \( \mathrm{O_2} \) are consumed, then \( 2x \) moles of \( \mathrm{NO} \) are produced.Initially, we have:- \( [N_2]_0 = 0.20 \text{ M} \)- \( [O_2]_0 = 0.15 \text{ M} \)- \( [NO]_0 = 0 \text{ M} \)At equilibrium:- \( [N_2] = 0.20 - x \)- \( [O_2] = 0.15 - x \)- \( [NO] = 2x \)Using stoichiometry and the equilibrium expression, we solve for \( x \) and find \( [NO] = 2x \). Ultimately, we calculate the equilibrium concentration of \( \mathrm{NO} \) to be 0.00348 M. Understanding the stoichiometric relationships is essential to determining these changes and solving the equilibrium problem accurately.