Problem 36 - Chemistry and Chemical Reactivity

Question

At \(2300 \mathrm{K}\) the equilibrium constant for the formation of \(\mathrm{NO}(\mathrm{g})\) is \(1.7 \times 10^{-3}\) $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g}) $$ (a) Analysis shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M},\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\) under certain conditions. Is the system at equilibrium? (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) When the system is at equilibrium, what are the equilibrium concentrations?

Step-by-Step Solution

Answer

(a) No, the system is not at equilibrium; (b) The reaction proceeds forward; (c) \([\mathrm{NO}] = 0.00672\,\text{M}, [\mathrm{N}_2] = [\mathrm{O}_2] = 0.2487\,\text{M}\) at equilibrium.

Step 1: Write the expression for the equilibrium constant

The equilibrium expression for the reaction \( \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2\mathrm{NO}(\mathrm{g}) \) is given by: \[ K_c = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_2][\mathrm{O}_2]} \] where \([\mathrm{NO}], [\mathrm{N}_2],\) and \([\mathrm{O}_2]\) are the molar concentrations of NO, \(\mathrm{N}_2,\) and \(\mathrm{O}_2\) respectively.

Step 2: Calculate the reaction quotient (Q)

The reaction quotient \(Q\) is calculated in the same manner as \(K_c\) but with the initial concentrations. Substitute the given concentrations into the expression: \[ Q = \frac{(0.0042)^2}{(0.25)(0.25)} = \frac{0.00001764}{0.0625} \approx 0.00028224 \]

Step 3: Compare Q and K_c

Compare the calculated \(Q\) with the given \(K_c\) (\(1.7 \times 10^{-3}\)). Since \(Q \approx 0.00028224 < 1.7 \times 10^{-3}\), the reaction is not at equilibrium, and it will shift towards the formation of products, NO.

Step 4: Determine the direction of the reaction

Since \(Q < K_c\), the reaction will proceed in the forward direction to reach equilibrium, meaning more \(\mathrm{NO}\) will be produced by consuming \(\mathrm{N}_2\) and \(\mathrm{O}_2\).

Step 5: Set up the expression for equilibrium concentrations

Let the change in concentration of \(\mathrm{NO}\) at equilibrium be \(+2x\). Then the changes for \(\mathrm{N}_2\) and \(\mathrm{O}_2\) are \(-x\). Set up the equation using the equilibrium constant: \[ K_c = \frac{(0.0042 + 2x)^2}{(0.25 - x)(0.25 - x)} = 1.7 \times 10^{-3} \]

Step 6: Solve for x

Since \(K_c\) is small, assume \( x \ll 0.25 \) and simplify: \[ K_c \approx \frac{(0.0042 + 2x)^2}{0.0625} \approx 1.7 \times 10^{-3} \] Solving for \(x\) gives: \[ x ≈ 0.00126 \text{ M} \]

Step 7: Calculate equilibrium concentrations

Using the value of \(x\), calculate the equilibrium concentrations: \([\mathrm{NO}] = 0.0042 + 2x = 0.0042 + 2(0.00126) = 0.00672\, \text{M}\). \([\mathrm{N}_2] = [\mathrm{O}_2] = 0.25 - x = 0.25 - 0.00126 = 0.2487\, \text{M}\).

Key Concepts

Chemical Equilibrium

Chemical equilibrium refers to the state of a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, although both reactions still occur, they do so at the same rate, resulting in no net change in the concentrations of reactants and products. Chemical equilibrium is a dynamic process, meaning that the particles constantly react but with no overall effect on the concentration levels.

In the case of the reaction \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftarrows 2\mathrm{NO},\) the system reaches equilibrium when the rate of formation of \(\mathrm{NO}\) equals its decomposition rate back into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). The equilibrium constant \(K_c\) provides a measure of this balanced state in terms of concentration, reflecting the ratio of concentrations of products to reactants at equilibrium.

The value of \(K_c\) can tell us much about the position of equilibrium. A large \(K_c\) suggests a greater concentration of products at equilibrium, indicating the reaction favors product formation. Conversely, a smaller \(K_c\) suggests the opposite.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a value that helps determine the position of a reaction relative to its equilibrium position. It is calculated using the same formula as the equilibrium constant \(K_c\), but with the concentrations at a particular moment in time, not necessarily at equilibrium.

For the reaction \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftarrows 2\mathrm{NO},\) the reaction quotient is given by:

• \( Q = \frac{[\mathrm{NO}]^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]}. \)

By comparing \(Q\) to \(K_c\), we can predict the direction in which the reaction will proceed to reach equilibrium:

• If \(Q < K_c\), the reaction will proceed forward, converting reactants to products to reach equilibrium.
• If \(Q > K_c\), the reaction will proceed in reverse, converting products back into reactants.
• If \(Q = K_c\), the system is already at equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle predicts how a system at equilibrium will respond to changes in concentration, temperature, or pressure. When a system at equilibrium is disturbed, it adjusts itself to minimize that disturbance and re-establish equilibrium.

For instance, in the reaction \(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftarrows 2\mathrm{NO},\) if additional \(\mathrm{N}_{2}\) or \(\mathrm{O}_{2}\) is added to the system, the equilibrium will shift in the direction that consumes these added reactants, i.e., towards the production of more \(\mathrm{NO}\).

• Adding reactants or removing products pushes the equilibrium towards more product formation (right shift).
• Removing reactants or adding products causes a shift towards more reactant formation (left shift).
• Changes in pressure or temperature can also cause shifts, depending on the specifics of the reaction.

Le Chatelier’s principle is a fundamental concept used to predict the behavior of a system when it experiences a change, thus guiding how one might manipulate reactions to favor the production of desired products or reactants.

Equilibrium Concentration

Equilibrium concentrations are the concentrations of reactants and products at the point of chemical equilibrium. These concentrations can be determined by using the equilibrium constant \(K_c\) in relation to the balanced chemical equation.

For the chemical system given:\(\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftarrows 2\mathrm{NO},\)the equilibrium concentrations can be calculated by setting up the equilibrium expression based on the stoichiometry of the reaction. Using the quadratic approximation method, if applicable, can simplify finding the equilibrium concentrations by assuming that changes in reactant concentrations are small relative to their initial amounts.

Steps for calculating equilibrium concentrations:

• Determine initial concentrations and calculate \(Q\) to understand the reaction direction.
• Write an expression for the equilibrium constant \(K_c\).
• Set up the equation based on the changes in concentration predicted by the reaction stoichiometry.
• Solve for the change in concentration to find the equilibrium concentrations of all species.

Knowing the equilibrium concentrations is crucial for predicting how a system will behave under different conditions and for practical applications in chemical manufacturing and other fields.