Problem 86

Question

The equilibrium constant \(K_{\mathrm{p}}\) for the thermal decomposition of \(\mathrm{NO}_{2}\) $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ is \(6.5 \times 10^{-6}\) at \(450^{\circ} \mathrm{C} .\) If a reaction vessel at this temperature initially contains only 0.500 atm \(\mathrm{NO}_{2},\) what will be the partial pressures of \(\mathrm{NO}_{2}, \mathrm{NO},\) and \(\mathrm{O}_{2}\) in the vessel when equilibrium has been attained?

Step-by-Step Solution

Verified
Answer
Answer: The equilibrium partial pressures of NO2, NO, and O2 are approximately 0.49562 atm, 0.00438 atm, and 0.00219 atm, respectively.
1Step 1: Write down the decomposition
The given decomposition reaction is: $$2 NO_{2}(g) \rightleftharpoons 2 NO(g) + O_{2}(g)$$
2Step 2: Set up an ICE table
An ICE table (Initial, Change, Equilibrium) helps organize the initial pressures, changes in pressures, and equilibrium pressures for each species in the reaction. 1 row for initial pressures (I): 0.500 atm NO2, 0 atm NO, and 0 atm O2 1 row for changes in pressures (C): -2x atm NO2, +2x atm NO, +x atm O2 1 row for equilibrium pressures (E): (0.500-2x) atm NO2, 2x atm NO, x atm O2 | | NO2 | 2NO | O2 | |--------|--------|-------|------| | I | 0.500 | 0 | 0 | | C | -2x | +2x | +x | | E | 0.500-2x| 2x | x |
3Step 3: Write the Kp expression
Using the balanced chemical equation and the equilibrium pressures, we can write the Kp expression: $$Kp = \frac{(P_{NO})^2(P_{O_2})}{(P_{NO_2})^2}$$
4Step 4: Substitute equilibrium pressures and given Kp
Substitute the equilibrium pressures from the ICE table and the given Kp value into the Kp expression: $$6.5 \times 10^{-6} = \frac{(2x)^2(x)}{(0.500-2x)^2}$$
5Step 5: Solve the equation for x
By solving this quadratic equation, we can find the value of x representing the change in pressures of each gas at equilibrium. The calculation might be difficult using analytical methods as the equation has an unknown higher than 1. Hence, we can solve it numerically or using a graphing calculator. The approximate value of x comes out to be 0.00219 atm.
6Step 6: Calculate the equilibrium pressures
Now that we know the value of x, we can find the equilibrium pressures for each gas using the ICE table: $$P_{NO_{2}} = 0.500 - 2(0.00219) = 0.49562\text{ atm}$$ $$P_{NO} = 2(0.00219) = 0.00438\text{ atm}$$ $$P_{O_{2}} = 0.00219\text{ atm}$$ When equilibrium is attained, the partial pressures of NO2, NO, and O2 are approximately 0.49562 atm, 0.00438 atm, and 0.00219 atm, respectively.

Key Concepts

Chemical EquilibriumICE TablePartial PressuresReaction QuotientThermal Decomposition
Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry where the rate of the forward reaction equals the rate of the backward reaction, and the concentrations of reactants and products remain constant over time. In the context of gaseous reactions, equilibrium involves the partial pressures of the reactants and products stabilizing to a particular ratio defined by the equilibrium constant, denoted as \( K_p \) when expressed in terms of partial pressures.

Understanding equilibrium is crucial because it explains why some reactions do not proceed to completion, and it helps predict the concentrations or partial pressures of the substances once the system has reached equilibrium. In the given exercise, we're dealing with the thermal decomposition of \( NO_2 \) gas, for which the equilibrium state is described by a constant value \( K_p \) that depends on temperature.
ICE Table
An ICE (Initial, Change, Equilibrium) table is an organized approach to solving equilibrium problems in chemistry. It structures the initial concentrations or partial pressures, changes due to the reaction, and the new equilibrium concentrations or partial pressures.

The use of ICE tables is particularly helpful when dealing with complex equilibrium systems, as it allows students to visually track the progression of the reaction from initial conditions to equilibrium. In the provided solution, the ICE table clearly outlines the changes in partial pressures of \( NO_2 \), \( NO \), and \( O_2 \) as the reaction proceeds toward equilibrium.
Partial Pressures
Partial pressures are the individual pressure components that each gas in a mixture contributes to the total pressure. Dalton's Law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.

In reactions involving gases, partial pressures are particularly important because they can affect reaction rates and equilibrium positions. The given exercise requires understanding of partial pressures in order to determine the equilibrium state of the system. The calculation of equilibrium partial pressures using the ICE table is a practical application of this concept.
Reaction Quotient
The reaction quotient, designated \( Q_p \), is a measure of the relative amounts of products and reactants present during a reaction at any given point in time, and is calculated in the same way as the equilibrium constant, but without the system necessarily being at equilibrium. It is useful for predicting which direction a reaction will proceed to reach equilibrium.

If \( Q_p < K_p \), the reaction will proceed in the forward direction to produce more products. Conversely, if \( Q_p > K_p \), the reaction will shift to produce more reactants. By comparing the reaction quotient to the known equilibrium constant, one can predict how the reaction will shift to achieve equilibrium.
Thermal Decomposition
Thermal decomposition is a chemical reaction where a compound breaks down into two or more substances when heated. It is often endothermic, meaning that the reaction absorbs energy. Thermal decomposition is a common method for preparing many compounds.

In the exercise, \( NO_2 \) decomposes into \( NO \) and \( O_2 \) upon heating. The reaction has an associated equilibrium constant that depends on temperature, reflecting the position of equilibrium at a given temperature. Understanding how thermal energy affects decomposition reactions is essential for predicting the behavior of reactants and products under heat.