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The reaction \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) has \(K_{\mathrm{P}}=0.140\) at \(25^{\circ} \mathrm{C}\). In a reaction vessel containing these gases in equilibrium at this temperature, the partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) was 0.250 atm. (a) What was the partial pressure of the \(\mathrm{NO}_{2}\) in the reaction mixture? (b) What was the total pressure of the mixture of gases?

Short Answer

Expert verified
The partial pressure of NO2 in the reaction mixture was approximately 0.210 atm, and the total pressure of the mixture of gases was approximately 0.670 atm.

Step by step solution

01

Write the Expression for the Equilibrium Constant Kp

The equilibrium constant, Kp, is expressed in terms of partial pressures of the gases involved in the equilibrium reaction. For the given reaction, it can be written as: \( K_{p} = \frac{(P_{NO_{2}})^2}{P_{N_{2}O_{4}}} \)
02

Substitute the Known Values into the Equation

Substitute the given values for Kp and the partial pressure of N2O4 into the equation to solve for the partial pressure of NO2. \( 0.140 = \frac{(P_{NO_{2}})^2}{0.250} \)
03

Solve for Partial Pressure of NO2

Isolate the term \((P_{NO_{2}})^2\) by multiplying both sides of the equation by the partial pressure of N2O4, and then take the square root of both sides to find PNO2. \( (P_{NO_{2}})^2 = 0.140 \times 0.250 \Rightarrow P_{NO_{2}} = \sqrt{0.140 \times 0.250} \)
04

Calculate the Total Pressure of the Gas Mixture

Since the reaction is at equilibrium, the total pressure of the gas mixture is the sum of the partial pressures of the individual gases. The total pressure P_total is given by: \( P_{total} = P_{N_{2}O_{4}} + 2 \times P_{NO_{2}} \)
05

Substitute the Known Values to Find Total Pressure

Substitute the known partial pressure of N2O4 and the calculated partial pressure of NO2 into the equation for P_total. \( P_{total} = 0.250 + 2 \times P_{NO_{2}} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant (Kp)
The equilibrium constant for a reaction involving gases, denoted as Kp, is a measure of the extent of the reaction when it reaches a state of balance, or equilibrium. At this point, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time.

For the reaction N2O4(g) ↔ 2NO2(g), we write the Kp expression by raising the partial pressures of the products to the power of their stoichiometric coefficients and dividing by the corresponding expression for the reactants:
\[\begin{equation}K_{p} = \frac{\left(P_{NO_{2}}\right)^2}{P_{N_{2}O_{4}}}\end{equation}\]
The value of Kp provided, 0.140, indicates the relative amounts of products and reactants at equilibrium at a given temperature, which in this case is 25oC.
Partial Pressure
Partial pressure represents the individual pressure exerted by a particular gas in a mixture of gases. It is directly proportional to its molar fraction in the mixture. The total pressure exerted by the gas mixture is the sum of the partial pressures of all gases present.

In our example, the partial pressure of N2O4 is given as 0.250 atm, and the partial pressure of NO2 can be calculated using the formula derived from the equilibrium constant expression. By determining the partial pressures of each component, one can also compute the total pressure of the gas mixture in the system.
Reaction Quotient
The reaction quotient, Q, is a measure that indicates the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, Kp, but with the initial partial pressures instead of the equilibrium partial pressures.

If Q > Kp, the reaction will proceed toward the reactants to reach equilibrium. Conversely, if Q < Kp, the reaction will proceed toward the products. When Q = Kp, the reaction is already at equilibrium. Calculating Q can predict shifts in the equilibrium position due to changes in concentration, pressure, or volume.
Le Chatelier's Principle
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system responds by shifting its equilibrium position in a way that counteracts the change. This can include changes in concentration, pressure, volume, or temperature.

For example, increasing the pressure of a gas reaction by decreasing volume generally shifts the equilibrium toward the side with fewer moles of gas. If temperature changes, the direction of the shift will favor the exothermic or endothermic reaction, depending on whether the temperature is decreased or increased respectively. Understanding Le Chatelier's principle helps predict how a reaction will respond to external changes and is crucial for controlling industrial chemical processes.

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Most popular questions from this chapter

At \(773^{\circ} \mathrm{C}\), a mixture of \(\mathrm{CO}(g), \mathrm{H}_{2}(g)\), and \(\mathrm{CH}_{3} \mathrm{OH}(g)\) was allowed to come to equilibrium. The concentrations were determined to be: \([\mathrm{CO}]=0.105 \mathrm{M},\left[\mathrm{H}_{2}\right]=0.250 \mathrm{M}\) \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.00261 M .\) Calculate \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$

Consider the following equilibrium. \(2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) If you were converting between \(K_{\mathrm{P}}\) and \(K_{\mathrm{c}}\), what value of \(\Delta n_{\mathrm{g}}\) would you use?

The reaction \(2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)\) has \(K_{\mathrm{c}}=\) \(3.2 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). If a reaction vessel contains initially \(0.0500 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{HCl}\) and then reacts to reach equilibrium, what will be the concentrations of \(\mathrm{H}_{2}\) and \(\mathrm{Cl}_{2}\) ?

Suppose for the reaction \(A \longrightarrow B\) the value of \(Q\) is less than \(K_{\mathrm{c}}\). Which way does the reaction have to proceed to reach equilibrium-in the forward or reverse direction?

At a certain temperature, \(K_{\mathrm{c}}=4.3 \times 10^{5}\) for the reaction $$ \mathrm{HCO}_{2} \mathrm{H}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ If \(0.200 \mathrm{~mol}\) of \(\mathrm{HCO}_{2} \mathrm{H}\) is placed in a \(1.00 \mathrm{~L}\) vessel, what will be the concentrations of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) when the system reaches equilibrium? (Hint: Where does the position of equilibrium lie when \(K\) is very large?)

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