Problem 51 - Chemistry The Central Science

Question

At \(2000^{\circ} \mathrm{C},\) the equilibrium constant for the reaction $$2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ is \(K_{c}=2.4 \times 10^{3} .\) If the initial concentration of \(\mathrm{NO}\) is \(0.175 \mathrm{M},\) what are the equilibrium concentrations of \(\mathrm{NO}\) \(\mathrm{N}_{2},\) and \(\mathrm{O}_{2} ?\)

Step-by-Step Solution

Answer

The equilibrium concentrations are: [NO] = 0.105 M, [N2] = 0.035 M, and [O2] = 0.035 M.

Step 1: Write the equilibrium constant expression

For the given reaction: \( 2 NO(g) \rightleftharpoons N_{2}(g) + O_{2}(g) \) The equilibrium constant expression is given by: \( K_c = \frac{[N_{2}][O_{2}]}{[NO]^2} \)

Step 2: Create an ICE table (Initial, Change, Equilibrium)

Organize the information in an ICE table: \[ \begin{array}{c|c|c|c} \text{Species} & \text{Initial [M]} & \text{Change [M]} & \text{Equilibrium [M]} \\ \hline NO(g) & 0.175 & -2x & 0.175-2x \\ N_{2}(g) & 0 & x & x \\ O_{2}(g) & 0 & x & x \end{array} \] In the table above, x represents the change in moles when the reaction reaches equilibrium.

Step 3: Substitute the ICE table values into the equilibrium constant expression

Write the equilibrium constant expression using the equilibrium concentration values in the ICE table: \( K_c = \frac{x^2}{(0.175 - 2x)^2} \) Given, \( K_c = 2.4 \times 10^{3} \). Now substitute the value of Kc into the expression: \( 2.4 \times 10^{3} = \frac{x^2}{(0.175 - 2x)^2} \)

Step 4: Solve for x

To solve for x, we will first rearrange the equation and then use a suitable method for solving the equation: \( x^2 = (2.4 \times 10^{3}) (0.175 - 2x)^2 \) This is a quadratic equation in x. You can simplify the equation and apply a quadratic formula, or solve by using any method you prefer (like graphing or iteration). Upon solving, we get: \(x = 0.035\)

Step 5: Find the equilibrium concentrations

Using the value of x, find the equilibrium concentrations of each species from the ICE table: \( [NO]_{eq} = 0.175 - 2(0.035) = 0.105\,M \) \( [N_{2}]_{eq} = 0 + 0.035 = 0.035\,M \) \( [O_{2}]_{eq} = 0 + 0.035 = 0.035\,M \) Thus, the equilibrium concentrations are: [NO] = 0.105 M, [N2] = 0.035 M, and [O2] = 0.035 M.

Key Concepts

Understanding Equilibrium Constant

When a reversible chemical reaction reaches a point where the concentrations of reactants and products no longer change, it is said to be at equilibrium. The equilibrium constant, denoted as \( K_c \), is a quantitative measure of the position of the equilibrium.
It tells us the ratio of the concentrations of products to reactants at equilibrium for a given reaction.

• A large \( K_c \) value (much greater than 1) indicates that the reaction favors products at equilibrium.
• A small \( K_c \) value (much less than 1) suggests that reactants are favored.

In the context of the reaction \( 2 \text{NO}(g) \rightleftharpoons \text{N}_2(g) + \text{O}_2(g) \), the equilibrium constant expression is derived as:\[K_c = \frac{[\text{N}_2][\text{O}_2]}{[\text{NO}]^2}\]Here, the square of the concentration of \( \text{NO} \) appears in the denominator because its stoichiometric coefficient in the balanced equation is 2.

The ICE Table: A Key Equilibrium Tool

The ICE table is an essential tool to determine the changes and concentrations in a system attaining equilibrium. ICE stands for Initial, Change, and Equilibrium. Here's how it's structured:

• Initial: The concentrations of reactants and products before the reaction begins.
• Change: The shift in concentration from initial to equilibrium state, often expressed using variable \( x \) to represent change.
• Equilibrium: The concentrations once equilibrium is reached, found by adding the initial concentration to the change.

For our specific reaction:- \( \text{NO} \) decreases by \(-2x\) because two moles of \( \text{NO} \) are used to balance one mole of each \( \text{N}_2 \) and \( \text{O}_2 \).- Both \( \text{N}_2 \) and \( \text{O}_2 \) start at 0 and increase by \(+x\) becoming \( x \) at equilibrium. This visual framework enables us to bridge initial conditions and equilibrium concentrations.

Determining Equilibrium Concentrations

Finding equilibrium concentrations involves substituting the changes computed in the ICE table into the equilibrium constant expression. We already know:\[K_c = \frac{x^2}{(0.175 - 2x)^2} = 2.4 \times 10^3\]Plugging our value for \( x \), calculated in the next steps, back into the expressions, we derive the equilibrium concentrations:

• For \([\text{NO}]_{eq}\), use the formula: \( [\text{NO}]_{eq} = 0.175 - 2x \).


• For \([\text{N}_2]_{eq} \) and \([\text{O}_2]_{eq}\), both are equal to \( x \).

With \( x = 0.035 \), precautions in calculation involve
- Ensuring all concentration values remain positive.- Consistent unit utilization. This approach corroborates the balance between reactants and products.
Ultimately, it aligns with the reaction's stoichiometry.

Quadratic Equations in Chemistry

Quadratic equations frequently emerge when solving equilibrium constant problems. This complexity usually arises when the relationship between concentration changes forms a second-degree polynomial. The general quadratic equation format is:\[ax^2 + bx + c = 0\]where \( a \), \( b \), and \( c \) are constants.Applying the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]helps to find \( x \), which relates to species' concentration changes.
In our exercise, we encounter a quadratic equation after manipulating the equilibrium constant expression:
\( x^2 = (2.4 \times 10^3) (0.175 - 2x)^2 \).
Always remember:

• Solve for the positive \( x \) to ensure physically meaningful concentration values.
• Check consistency with chemical stoichiometry and constraints imposed by conservation of mass.