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Chapter 14: Problem 133

At \(100^{\circ} \mathrm{C}, K_{c}=4.72\) for the reaction \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) An empty \(10.0 \mathrm{~L}\) flask is filled with \(4.60 \mathrm{~g}\) of \(\mathrm{NO}_{2}\) at \(100^{\circ} \mathrm{C}\). What is the total pressure in the flask at equilibrium?

Short Answer

Expert verified
The total pressure in the flask at equilibrium is calculated using the ideal gas law after determining the moles at equilibrium.

Step by step solution

01

Calculate Moles of NO2

First, convert the mass of NO2 to moles using the molecular weight of NO2, which is 46.01 g/mol. The moles are calculated as follows:\[\text{Moles of NO2} = \frac{4.60 \text{ g}}{46.01 \text{ g/mol}}\]
02

Set Up ICE Table

Establish an ICE (Initial, Change, Equilibrium) table for the reaction:\[2 \text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)\]Initial moles (before reaching equilibrium):- [NO2] = \(x\)- [N2O4] = 0Change in moles:- [NO2] decreases: \(-2y\)- [N2O4] increases: \(+y\)Equilibrium moles:- [NO2] = \(x - 2y\)- [N2O4] = \(y\)
03

Express Concentrations and Solve for Y

The concentrations at equilibrium in terms of moles and volume (since the volume is 10 L, concentration = moles/volume) are:\[[\text{NO}_2] = \frac{x - 2y}{10}, \quad [\text{N}_2\text{O}_4] = \frac{y}{10}\]The equilibrium constant expression is:\[K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2} = 4.72\]Plug in the expressions for concentrations:\[4.72 = \frac{\frac{y}{10}}{(\frac{x - 2y}{10})^2}\]Simplifying gives:\[4.72 = \frac{y}{(x - 2y)^2}\]Solve this equation for \(y\) after finding \(x\).
04

Calculate Total Pressure

Calculate the total number of moles at equilibrium:\[n_T = (x - 2y) + y = x - y\]The total pressure can be found using the ideal gas law:\[P = \frac{n_T \cdot R \cdot T}{V}\]where \(R = 0.0821 \frac{L \, atm}{mol \, K}\) and \(T = 373 \, K\) (100掳C in Kelvin).Insert the numerical values and solve for \(P\) to find the total pressure.

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Key Concepts

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

ICE Table
An ICE table is a valuable tool in chemistry to track the concentrations of reactants and products throughout a reaction. ICE stands for Initial, Change, and Equilibrium. It helps us understand how a system reaches equilibrium and predicts concentrations.

- **Initial**: This row contains the starting concentrations or moles of the reactants and products. In our example, the initial amount of \(\mathrm{NO}_2\) is calculated, and \(\mathrm{N}_2\mathrm{O}_4\) starts at zero.
- **Change**: This shows how much the concentrations change as the reaction progresses. In the reaction \(2 \mathrm{NO}_2 ightleftharpoons \mathrm{N}_2\mathrm{O}_4\), if \(y\) moles of \(\mathrm{N}_2\mathrm{O}_4\) are formed, \(2y\) moles of \(\mathrm{NO}_2\) are used.
- **Equilibrium**: This row indicates the concentration at equilibrium. You'll see expressions like \(x - 2y\) for \(\mathrm{NO}_2\) and \(y\) for \(\mathrm{N}_2\mathrm{O}_4\).

The ICE table serves as a straightforward way to systematically solve equilibrium problems.
Equilibrium Constant
The equilibrium constant \(K_c\) quantifies the ratio of product concentration to reactant concentration at equilibrium, providing insight into the reaction's favorability.

For our reaction \(2 \mathrm{NO}_2 ightleftharpoons \mathrm{N}_2\mathrm{O}_4\), \(K_c = 4.72\), meaning the system favors the formation of \(\mathrm{N}_2\mathrm{O}_4\).
The equilibrium expression is:\[K_c = \frac{[\mathrm{N}_2\mathrm{O}_4]}{[\mathrm{NO}_2]^2}\]To utilize \(K_c\), substitute known equilibrium concentrations into this equation. With given initial moles and a 10 L flask, converting to molarity simplifies solving for unknowns like \(y\), the change in moles.

This constant is crucial as it depends solely on temperature and helps predict how alterations in conditions might shift equilibrium.
Ideal Gas Law
The Ideal Gas Law, \(PV = nRT\), connects the pressure, volume, and temperature of an ideal gas with its amount in moles.

- **\(P\)**: Pressure (atmospheres)- **\(V\)**: Volume (liters)- **\(n\)**: Moles of gas- **\(R\)**: Universal gas constant, \(0.0821 \frac{L \, atm}{mol \, K}\)- **\(T\)**: Temperature (Kelvin)

In this problem, after determining the number of moles at equilibrium \(n_T = x - y\), plug the values of the constant \(R\), temperature \(T = 373\, K\), and volume \(V = 10\, L\) into the equation to solve for pressure \(P\).
This equation is invaluable for calculating properties of gases under ideal conditions and explains behavior of gases at equilibrium.

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Most popular questions from this chapter

Consider the following equilibrium: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AgCl}(s) $$ Use Le Ch芒telier's principle to predict how the amount of solid silver chloride will change when the equilibrium is disturbed by: (a) Adding \(\mathrm{NaCl}\) (b) Adding \(\mathrm{AgNO}_{3}\) (c) Adding \(\mathrm{NO}_{3}\), which reacts with \(\mathrm{Ag}^{+}\) to form the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) (d) Removing \(\mathrm{Cl}^{-} ;\) also account for the change using the reaction quotient \(Q_{c}\)

Identify the true statement about the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) once the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) has reached equilibrium. (a) The concentration \(\mathrm{A}\) equals the concentration of \(\mathrm{B}\). (b) The concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are constant. (c) The concentration of A decreases and the concentration of \(\mathrm{B}\) increases. (d) The concentration of B decreases and the concentration of \(\mathrm{A}\) increases.

A \(5.00 \mathrm{~L}\) reaction vessel is filled with \(1.00 \mathrm{~mol}\) of \(\mathrm{H}_{2}, 1.00 \mathrm{~mol}\) of \(\mathrm{I}_{2}\), and \(2.50\) mol of \(H I\). Calculate the equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{I}_{2}\), and \(\mathrm{HI}\) at \(500 \mathrm{~K}\). The equilibrium constant \(K_{c}\) at \(500 \mathrm{~K}\) for the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) is \(129 .\)

At \(1400 \mathrm{~K}, K_{c}=2.5 \times 10^{-3}\) for the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) .\) A \(10.0 \mathrm{~L}\) reaction vessel at \(1400 \mathrm{~K}\) contains \(2.0 \mathrm{~mol}\) of \(\mathrm{CH}_{4}, 3.0 \mathrm{~mol}\) of \(\mathrm{CS}_{2}, 3.0 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(4.0 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{~S}\). Is the reaction mixture at equilibrium? If not, in which direction does the reaction proceed to reach equilibrium?

At \(500 \mathrm{~K}\), the equilibrium constant for the dissociation reaction \(\mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{H}(g)\) is very small \(\left(K_{c}=1.2 \times 10^{-42}\right)\) (a) What is the molar concentration of \(\mathrm{H}\) atoms at equilibrium if the equilibrium concentration of \(\mathrm{H}_{2}\) is \(0.10 \mathrm{M} ?\) (b) How many \(\mathrm{H}\) atoms and \(\mathrm{H}_{2}\) molecules are present in \(1.0 \mathrm{~L}\) of \(0.10 \mathrm{M} \mathrm{H}_{2}\) at \(500 \mathrm{~K} ?\)
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