Question

At \(2200^{\circ} \mathrm{C}, K_{\mathrm{p}}=0.050\) for the reaction $$ \mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ What is the partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that were placed in a flask at initial pressures of \(0.80\) and \(0.20\) atm, respectively?

Short answer

The partial pressure of \(\mathrm{NO}\) in equilibrium with \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is approximately \(0.204\, \text{atm}\).

Step-by-step solution

Write the reaction and equilibrium expression

Write down the given reaction and the equilibrium expression for Kp: \(\mathrm{N}_{2} (\mathrm{g}) + \mathrm{O}_{2} (\mathrm{g}) \rightleftharpoons 2 \mathrm{NO} (\mathrm{g})\) \( K_{\mathrm{p}} = \frac{\text{[NO]}^2}{[\mathrm{N}_{2}][\mathrm{O}_{2}]} \)

Define the changes in partial pressures

Since we are given the initial pressures of N2 and O2, let's define the changes in their partial pressures with \(x\): Change in N2 pressure: -x Change in O2 pressure: -x Change in NO pressure: +2x

Write the expressions for the equilibrium partial pressures

Incorporate the changes in partial pressures into the equilibrium expression of each species: \([\mathrm{N}_2]_{eq} = 0.80-x\) \([\mathrm{O}_2]_{eq} = 0.20-x\) \([\mathrm{NO}]_{eq} = 2x\)

Substitute the equilibrium expressions into the Kp equation

Substitute the expressions for the equilibrium partial pressures into the Kp equation given: \( 0.050 = \frac{(2x)^2}{(0.80-x)(0.20-x)} \)

Solve for x

Solve the equation above for x: Solving the equation algebraically or numerically, we get: \(x \approx 0.102\)

Calculate equilibrium partial pressure of NO

Finally, calculate the equilibrium partial pressure of NO using the value of x: \([\mathrm{NO}]_{eq} = 2x = 2(0.102) \approx 0.204\, \text{atm}\) The partial pressure of NO in equilibrium with N2 and O2 is approximately 0.204 atm.

Key concepts

Equilibrium Constant

In chemical reactions, the equilibrium constant, often represented as \( K_p \) or \( K_c \), is a crucial factor that helps us understand the balance between reactants and products at equilibrium. For gaseous reactions like the one involving nitrogen (\( \mathrm{N}_2 \)), oxygen (\( \mathrm{O}_2 \)), and nitrogen monoxide (\( \mathrm{NO} \)), \( K_p \) specifically relates to partial pressures.

The equilibrium constant formula is constructed by dividing the products' partial pressures, each raised to the power of their coefficients, by the reactants' partial pressures, similarly adjusted. In the provided example, the equilibrium constant \( K_p = 0.050 \), given by:

• Numerator: \([\mathrm{NO}]^2\) because \(2\mathrm{NO} \) shows it forms 2 moles for every mole of reactants.
• Denominator: \([\mathrm{N}_2][\mathrm{O}_2]\) where one mole of each reactant is involved in the initial reaction.

Such constants are temperature dependent; therefore, this \( K_p \) value applies specifically at \( 2200^{\circ} \mathrm{C} \).

Partial Pressure

Partial pressure is an essential concept when dealing with gaseous equilibria. It refers to the pressure that a single gas in a mixture would exert if it occupied the entire volume on its own. In the example problem, initial partial pressures are given as \(0.80\) atm for \(\mathrm{N}_2\) and \(0.20\) atm for \(\mathrm{O}_2\).

• The change in these pressures as the reaction progresses is critical when establishing equilibrium.
• Partial pressures at equilibrium are determined by taking the initial pressures and incorporating the change due to reaction progress, often denoted by \(x\).

For \( \mathrm{N}_2 \) and \( \mathrm{O}_2 \), the pressures decrease by \(x\) whereas \( \mathrm{NO} \), being a product, increases by \(2x\). Calculating \(x\) allows us to find the exact equilibrium pressures, with \([\mathrm{NO}]_{eq} = 2x = 0.204\) atm in the solution.

Gaseous Reaction

The reaction under study, \(\mathrm{N}_2 (\mathrm{g}) + \mathrm{O}_2 (\mathrm{g}) \rightleftharpoons 2 \mathrm{NO} (\mathrm{g})\), is a classic example of a gaseous reaction. Such reactions involve all reactants and products in the gaseous state. This uniform phase can simplify calculations, as everything can be expressed in terms of partial pressures.

• Gaseous reactions are particularly relevant because they involve volume and pressure, which are interdependent in gases.
• In this reaction, the balance shifts from reactants \(\mathrm{N}_2\) and \(\mathrm{O}_2\) to the product \(\mathrm{NO}\) as equilibrium is established.

The dynamic nature of gaseous reactions means that tiny changes in conditions, like pressure or temperature, can significantly impact the position of equilibrium and the amounts of products and reactants.

Le Chatelier's Principle

Le Chatelier's Principle is a valuable concept in understanding how systems at equilibrium respond to external changes. It predicts how the equilibrium will shift when conditions change, such as pressure, temperature, or concentrations.

• If a change in pressure occurs by adding or removing a gaseous reactant or product, the equilibrium will adjust to oppose this change.
• The principle suggests that adding more \(\mathrm{N}_2\) or \(\mathrm{O}_2\) would shift the equilibrium to favor the production of more \(\mathrm{NO} \), aiming to re-establish balance by increasing the forward reaction rate.

In the context of the given problem, understanding this principle helps interpret potential changes in the reaction conditions and their effect on the equilibrium position. Le Chatelier's Principle is a powerful tool for predicting the direction of change in reversible reactions and ensuring desired products.