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 \(\mathrm{CO}_{2}\) ?

Short Answer

Expert verified
The partial pressure of \(\text{CH}_4\) increases when \(\text{CO}_2\) partial pressure is increased (a) and also increases when \(\text{CH}_4\) partial pressure is decreased (b). The equilibrium constant remains unchanged with changes in \(\text{CH}_4\) concentration (c). The concentration of \(\text{CO}_2\) will increase if the concentration of \(\text{H}_2 \text{O}\) is decreased (d).

Step by step solution

01

Identify the Reaction and Direction of Shift

The given reaction is \(\text{CH}_4(g) + 2 \text{O}_2(g) = \text{CO}_2(g) + 2 \text{H}_2\text{O}(g)\). This is an equilibrium reaction, and any change in concentration, pressure, volume, or temperature can shift the reaction left or right according to Le Chatelier's Principle.
02

Analyze the Effect of Increasing \(\text{CO}_2\) Partial Pressure on \(\text{CH}_4\)

According to Le Châtelier’s Principle, increasing the partial pressure of \(\text{CO}_2\) will cause the equilibrium to shift to the left to reduce the effect of increased \(\text{CO}_2\). Consequently, the partial pressure of \(\text{CH}_4\) will increase.
03

Analyze the Effect of Decreasing \(\text{CH}_4\) Partial Pressure on \(\text{CO}_2\)

According to Le Châtelier’s Principle, decreasing the partial pressure of \(\text{CH}_4\) will cause the equilibrium to shift to the right to oppose the reduction of \(\text{CH}_4\). As a result, the partial pressure of \(\text{CO}_2\) will increase.
04

Analyze the Effect of Increasing \(\text{CH}_4\) Concentration on Equilibrium Constant

The equilibrium constant (\text{K}) for a reaction at a given temperature is constant and does not change with concentrations of reactants or products. Therefore, increasing the concentration of \(\text{CH}_4\) will shift the position of the equilibrium but will not affect the value of \text{K}.
05

Analyze the Effect of Decreasing \(\text{H}_2 \text{O}\) Concentration on \(\text{CO}_2\) Concentration

According to Le Châtelier’s Principle, decreasing the concentration of \(\text{H}_2 \text{O}\) will cause the equilibrium to shift to the right to oppose the decrease in \(\text{H}_2 \text{O}\). This will lead to an increase in the concentration of \(\text{CO}_2\).

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Key Concepts

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

Chemical Equilibrium
In chemistry, chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of the reactants and products remain constant over time as they are converted into each other. It's important to note that this doesn't mean the reactants and products are in equal concentrations, but that their ratios do not change.

Equilibrium is dynamic, meaning the reactions continue to occur, but there is no net change in the concentration of the chemicals involved. Think of it as a busy crosswalk where the same number of people enter and leave, maintaining a consistent overall crowd. In the context of the given exercise, the reaction between methane and oxygen to form carbon dioxide and water reaches such a balanced state. However, changing conditions, such as pressure or concentration, can disrupt this balance, causing the system to shift in order to restore equilibrium, as described by Le Chatelier's Principle.
Partial Pressure
The term partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. It's directly proportional to the concentration of the gas in the mixture. This concept is crucial when dealing with gases, as each gas in a mixture behaves independently and contributes to the total pressure of the mixture in accordance to Dalton's Law of Partial Pressures.

When discussing chemical equilibrium in gaseous systems, changes in partial pressures can be used as a tool to understand shifts in equilibrium.

Understand the Pressure-Impact

In our exercise, the increase of carbon dioxide's partial pressure indicates a higher quantity of CO2 present, triggering the system to adjust by increasing the production of reactants—like a seesaw balancing itself when one side becomes heavier.
Equilibrium Constant
The equilibrium constant (K) is a value that expresses the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction at a given temperature. This constant is solely dependent on temperature and is indispensable for predicting the direction of equilibrium shifts.

Every reversible chemical reaction has its own unique K value under a specific set of conditions. The magnitude of K provides insight into the position of equilibrium; a larger K suggests a product-favored reaction, while a smaller K indicates a reactant-favored reaction.

No Change in Constants

In the context of the problem presented, it is crucial to note that changing the concentration of methane (CH4) does not alter the K value, even though it leads to a shift in the equilibrium position to oppose this change—the 'constant' in equilibrium constant is there for a reason.
Reaction Concentration Changes
When we talk about reaction concentration changes, we are looking at the impact of altering the amounts of reactants or products in a chemical equilibrium. According to Le Chatelier's Principle, a system at equilibrium will respond to these changes in a way that 'opposes' the change, essentially attempting to return to its original state of balance.

In the exercise we're examining, increasing the concentration of methane (CH4) will result in a shift toward the formation of more products, in this case, CO2 and H2O, to reduce the impact of added CH4. Conversely, decreasing the concentration of a product, such as water (H2O), would shift in the opposite direction, increasing the production of reactants to restore balance.

These adjustments illustrate the incredible 'self-balancing' nature of chemical systems and the underlying predictability they exhibit in response to external changes.

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Most popular questions from this chapter

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

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Fe}(\mathrm{OH})_{3}(\mathrm{~s}) \rightarrow \mathrm{Fe}^{3}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq})\) (b) \(\mathrm{CuSO}_{4}(\mathrm{~s})+5 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) (s) (c) \(\mathrm{ZnO}(\mathrm{s})+\mathrm{CO}(\mathrm{g}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})\)

(a) Calculate the reaction free energy of \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) when the partial pressures of \(\mathrm{N}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NH}_{3}\) are \(1.0,4.2\), and 63 bar, respectively, and the temperarure is \(400 \mathrm{~K}\). For this reaction, \(K=41\) at \(400 \mathrm{~K}\). (b) Indicate whether this reaction mixture is likely to form reactants, is likely to form products, or is at equilibrium.

A mixture consisting of \(1.000 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) and \(1.000 \mathrm{~mol} \mathrm{CO}(\mathrm{g})\) is placed in a \(10.00-\mathrm{L}\) reaction vessel at \(800 \mathrm{~K}\). At equilibrium, \(0.665 \mathrm{~mol} \mathrm{CO}_{2}(\mathrm{~g})\) is present as a result of the reaction \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\) \(\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\). What are (a) the equilibrium concentrations for all substances and (b) the value of \(K_{e}\) ?

\( \mathrm{~A} 25.0 \mathrm{~g}\) sample of ammonium carbamate, \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), was placed in an evacuated \(0.250-\mathrm{L}\). flask and kept at \(25^{\circ} \mathrm{C}\). At equilibrium, the flask contained \(17.4 \mathrm{mg}\) of \(\mathrm{CO}_{2}\). What is the value of \(K_{c}\) for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s})=2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

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