Problem 21 - Chemistry The Central Science In Si Units

Question

If \(K_{c}=0.013 \mathrm{~L} / \mathrm{mol}\) for \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) at \(1000 \mathrm{~K},\) what is the value of \(K_{p}\) for this reaction at this temperature?

Step-by-Step Solution

Answer

The value of \(K_p\) is approximately 0.000159.

Step 1: Find Reaction Type and Given Formula

This is a chemical equilibrium problem involving gases. We are given the equilibrium constant in terms of concentration, \(K_c = 0.013 \text{ L/mol}\), for the reaction \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) at temperature \(1000 \text{ K}\).

Step 2: Identify the Relation between Kc and Kp

The relationship between \(K_c\) and \(K_p\) for gaseous reactions is given by the formula: \[ K_p = K_c (RT)^{\Delta n}\] where \(R\) is the gas constant \(0.08206 \, \text{L atm/mol K}\), and \(\Delta n\) is the change in moles of gas from reactants to products.

Step 3: Calculate \(\Delta n\)

\(\Delta n\) is the change in the number of moles of gas between the products and reactants. Here, the balanced equation is \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\). Total moles of gaseous products are 2, and moles of gaseous reactants are 2 + 1 = 3. Therefore, \(\Delta n = 2 - 3 = -1\).

Step 4: Calculate Kp Using the Formula

Substitute the values into the formula \(K_p = K_c (RT)^{\Delta n}\). Therefore: \[K_p = 0.013 \times (0.08206 \times 1000)^{-1}\] Calculate \((0.08206 \times 1000) = 82.06\), hence: \[K_p = 0.013 \times (82.06)^{-1}\] \[K_p = 0.013 \times \frac{1}{82.06}\] \[K_p \approx 0.013 \times 0.0122 \approx 0.0001586\]

Step 5: Finalize the Answer

Round the calculated \(K_p\) to three significant figures as typically needed in chemistry. \[K_p \approx 0.000159\]

Key Concepts

Kc and Kp relationship

In chemical equilibrium, particularly involving gaseous reactions, both the equilibrium constants, \(K_c\) and \(K_p\), play crucial roles. \(K_c\) is expressed in terms of the concentrations of reactants and products, making it more useful when studying reactions in solutions. Meanwhile, \(K_p\) is used when dealing with gaseous reactions, as it is expressed in terms of partial pressures.

The relationship between \(K_c\) and \(K_p\) is captured by the equation: \[ K_p = K_c (RT)^{\Delta n} \] Here:

• \(R\) represents the universal gas constant, valued at \(0.08206 \text{ L atm/mol K}\).
• \(T\) is the absolute temperature measured in Kelvin.
• \(\Delta n\) stands for the change in moles of gas in the reaction.

This formula highlights how changes in pressure can influence the equilibrium constant in gaseous systems, thus affecting the position of equilibrium under varying conditions.

gaseous reactions

Gaseous reactions are characterized by their involvement of gaseous reactants and products. These reactions can vary from the simple combination of gases to form new compounds, to more complex mechanisms that involve multiple steps.

In such reactions, the measurement of concentrations via pressures is more appropriate, as gases are defined by their volumes and pressures. Using partial pressures instead of molar concentration simplifies the calculations and provides a clear insight into the behavior of gas particles at equilibrium. Working with \(K_p\) in these scenarios gives a more direct understanding, reflecting the influence of temperature and pressure changes on equilibrium.
For example, in the reaction given: \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\), the starting and resulting gases and their pressures control the equilibrium position.

equilibrium constant calculation

Calculating the equilibrium constant involves determining the ratio of the concentration or pressure of products to reactants, at equilibrium, each raised to the power of their coefficients in the balanced chemical equation.
For reactions with gas, determining \(K_p\) requires knowing \(K_c\) and the conditions under which the reaction takes place, specifically temperature and mole changes. With \(K_p\), we bring in the equation:
\[ K_p = K_c (RT)^{\Delta n} \] Substitute known values, including the gas constant \(R\) and temperature \(T\), to derive \(K_p\). This results in an equilibrium constant that directly correlates with pressure changes within the system.
Calculations such as:

• \(K_p = 0.013 \times (82.06)^{-1}\)
• \(K_p \approx 0.013 \times 0.0122 \approx 0.0001586\)

help in understanding how pressure dynamics influence equilibrium in gases.

change in moles, \(\Delta n\)

In the context of the relationship between \(K_c\) and \(K_p\), \(\Delta n\) is vital as it represents the change in moles of gases when going from the reactants to the products in a balanced chemical equation.

For a shift in equilibrium from reactants to products:

• Calculate the total moles of gaseous products.
• Subtract the total moles of gaseous reactants from this value.
• The result is \(\Delta n\).

In the provided reaction, \(2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\), the gaseous products have 2 moles whereas the gaseous reactants have 3 moles (2 from \(\mathrm{NO}\) and 1 from \(\mathrm{Br}_2\)). Thus, the calculation of \(\Delta n = 2 - 3 = -1\) is necessary to accurately compute the relationship between \(K_c\) and \(K_p\).
This change in moles indicates how the system responds to pressure and temperature variations, which is pivotal for equilibria shifting and calculation accuracy.