Explain why the following cquilibria are heterogeneous and write the reaction quotient \(Q\) for each one. (a) \(2 \mathrm{Fe}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{FcCl}_{3}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g})\) (c) \(2 \mathrm{KNO}_{3}\) (s) \(\rightleftharpoons 2 \mathrm{KNO}_{2}\) (s) \(+\mathrm{O}_{2}\) (g)

Short Answer

Expert verified
The equilibria are heterogeneous because they include species in different phases: (a) and (c) include both solid and gas, while (b) involves a solid and two gases. The reaction quotient Q for each is (a) \( Q = \frac{1}{[\mathrm{Cl}_2]^3} \), (b) \( Q = [\mathrm{NH}_3]^2 \cdot [\mathrm{CO}_2] \), and (c) \( Q = [\mathrm{O}_2] \).

Step by step solution

01

Define a heterogeneous equilibrium

A heterogeneous equilibrium involves reactants and products in different phases, such as solid, liquid, and gas. To identify if an equilibrium is heterogeneous, look for the presence of different states of matter among the reactants and products.
02

Identify the states of matter in the given equilibria

For each given equilibrium, identify the physical state of each substance using the information provided in the parentheses: (s) for solid, (l) for liquid, and (g) for gas.
03

Justify the heterogeneity for each equilibrium reaction

Analyze the physical states of reactants and products for each reaction to determine if they are different. If they are different, then the equilibrium is heterogeneous.
04

Write the reaction quotient Q for (a)

The reaction quotient, Q, takes into account the concentrations of the gaseous reactants and products, and the partial pressures for gases. Pure solids and pure liquids do not appear in the expression for Q. So for reaction (a), Q is determined only by the gaseous reactants: \( Q = \frac{1}{[\mathrm{Cl}_2]^3} \), since \( \mathrm{Fe}(s) \) and \( \mathrm{FeCl}_3(s) \) are not included.
05

Write the reaction quotient Q for (b)

For reaction (b), Q is determined by the gaseous products: \( Q = \frac{[\mathrm{NH}_3]^2 \cdot [\mathrm{CO}_2]}{1} \), since \( \mathrm{NH}_4(\mathrm{NH}_2 \mathrm{CO}_2)(s) \) is a pure solid and does not appear in the expression.
06

Write the reaction quotient Q for (c)

For reaction (c), Q is determined by the gaseous product: \( Q = \frac{[\mathrm{O}_2]}{1} \), while the solids \( 2 \mathrm{KNO}_3(s) \) and \(2 \mathrm{KNO}_2(s) \) do not appear in Q.

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Key Concepts

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

Reaction Quotient Q
To determine how close a chemical reaction in a state of non-equilibrium is to reaching equilibrium, scientists use a helpful tool called the reaction quotient, denoted as 'Q'. It considers the concentrations (or partial pressures) of reactants and products at any point in time during the reaction.

For a reaction involving gases, Q is calculated using the formula:
\[ Q = \frac{\text{Product Concentrations}}{\text{Reactant Concentrations}} \]
This formula is adjusted to exclude solids and liquids, since their concentrations don't change during the reaction. If the reaction includes solids or pure liquids, they are essentially ignored in the Q expression, leaving only gases — and occasionally aqueous solutions — to consider.

For example, in the equilibrium (a) from the exercise, \( Q = \frac{1}{[\mathrm{Cl}_2]^3} \), the solid iron (Fe) and iron(III) chloride (FeCl3) are not included. This is different from the equilibrium constant 'K', where K equals Q only when the system is at equilibrium. Ultimately, Q tells us the direction in which the reaction must shift to reach equilibrium, providing a snapshot of the system's progress.
Physical States of Matter
Understanding the physical states of matter is essential when analyzing chemical reactions, especially those occurring in heterogeneous equilibria. Matter exists principally in three states—solid, liquid, and gas—which are represented by 's', 'l', and 'g' respectively in chemical equations.

Solids have a fixed volume and shape, liquids adapt to the shape of their container while maintaining a constant volume, and gases fill the entirety of their container regardless of volume. These properties profoundly impact how substances react chemically and physically with each other.

In our given exercise, the diverse states of matter present in the reactions define them as heterogeneous equilibria. For instance, in the exercise's reaction (b), we have a solid ammonium carbamate decomposing into two gases, ammonia, and carbon dioxide. The shift in physical states from solid to gas is a key characteristic that defines this reaction's heterogeneity.
Chemical Equilibrium
Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate, resulting in no overall change in the amounts of reactants and products over time. It is a dynamic state, meaning the reactions continue to occur but balance each other out.

At the macroscopic level, everything appears still, but on the microscopic level, reactants are constantly converting to products and vice versa. This balancing act can be influenced by changes in concentration, pressure, temperature, or the presence of catalysts.

Understanding equilibrium is essential when predicting how a reaction will respond to such changes, which is described by Le Chatelier's Principle. It is important to note that the concept of equilibrium applies to both homogeneous and heterogeneous reactions. However, in a heterogeneous equilibrium, like the ones in our exercise, the solid phases do not shift or 'react', per se; instead, they provide a constant surface area and are therefore omitted from the equilibrium expression.

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Most popular questions from this chapter

At \(1565 \mathrm{~K}\), the equilibrium constants for the reactions (1) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) and \((2)\) \(2 \mathrm{CO}_{2}(\mathrm{~g})=2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(1.6 \times 10^{-11}\) and \(1.3 \times 10^{-10}\), respectively. (a) What is the equilibrium constant for the reaction (3) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})=\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) at that temperature? (b) Show that the manner in which equilibrium constants are calculated is consistent with the manner in which the \(\Delta G_{1}^{\circ}\) values are calculated when combining two or more equarions by determining \(\Delta G_{e}{ }^{\circ}\) for \((1)\) and \((2)\) and using those values to calculare \(\Delta G,{ }^{\circ}\) and \(K_{3}\) for reaction (3).

Consider the equilibrium \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) If the partial pressure of \(\mathrm{CO}_{2}\) is increased, what happens to the partial pressure of \(\mathrm{CH}_{4}\) ? (b) If the partial pressure of \(\mathrm{CH}_{4}\) is decreased, what happens to the partial pressure of \(\mathrm{CO}_{2}\) ? (c) If the concentration of \(\mathrm{CH}_{4}\) is increased, what happens to the equilibrium constant for the reaction? (d) If the concentration of \(\mathrm{H}_{2} \mathrm{O}\) is decreased, what happens to the concentration of \(\mathrm{CO}_{2}\) ?

A reaction mixture consisting of \(2.00 \mathrm{~mol} \mathrm{CO}\) and \(3.00 \mathrm{~mol} \mathrm{H}_{2}\) is placed into a \(10.0-\mathrm{L}\) reaction vessel and heated to \(1200 \mathrm{~K}\). At cquilibrium, \(0.478 \mathrm{~mol} \mathrm{CH}_{4}\) was present in the system. Determine the value of \(K_{c}\) for the reaction \(\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{~g})+\) \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\).

At \(500^{\circ} \mathrm{C}, K_{c}=0.061\) for \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons\) \(2 \mathrm{NH}_{3}(\mathrm{~g})\). If analysis shows that the composition is \(3.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~N}_{2}, 2.00 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{H}_{2}\) and \(0.500 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{NH}_{3}\), is the reaction at equilibrium? If not, in which direction does the reaction tend to proceed to reach equilibrium?

Determine \(K_{c}\) from the value of \(K\) for cach of the following equilibria. (a) \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{3}(\mathrm{~g}), K=3.4\) at \(1000 \mathrm{~K}\) (b) \(\mathrm{NH}_{4} \mathrm{HS}(\mathrm{s})=\mathrm{NH}_{3}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}), K=9.4 \times 10^{-2}\) at \(24^{\circ} \mathrm{C}\)

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