Consider the equilibrium \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g})=\) \(4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\). (a) What happens to the partial pressure of \(\mathrm{NH}_{3}\) when the partial pressure of \(\mathrm{NO}\) is increased? (b) Does the partial pressure of \(\mathrm{O}_{2}\) decrease when the partial pressure of \(\mathrm{NH}_{3}\) decreases?

Short Answer

Expert verified
When the partial pressure of NO is increased, the partial pressure of NH3 increases; when the partial pressure of NH3 decreases, the partial pressure of O2 also decreases.

Step by step solution

01

Understand the Problem and Principle

We are to analyze the effects of changing partial pressures on a balanced chemical equation at equilibrium. The principle guiding this analysis is Le Chatelier's Principle, which states that if an external change is imposed on a system at equilibrium, the system adjusts itself to partially oppose the change and to establish a new equilibrium.
02

Assess the Increase of NO Partial Pressure

According to the given equilibrium, 4 molecules of NH3 react with 5 molecules of O2 to form 4 molecules of NO and 6 molecules of H2O, all in their gaseous state. Applying Le Chatelier's Principle, if we increase the partial pressure of NO, the system will attempt to reduce this stress by favoring the reaction that consumes NO — the reverse reaction. Therefore, the equilibrium shifts to the left, and as a result, the partial pressure of NH3 will increase.
03

Evaluate Decrease in NH3 Partial Pressure

Similarly, if the partial pressure of NH3 decreases, the system again will respond by shifting the equilibrium to reduce that change, which means the forward reaction will be favored. As NH3 is consumed, the reaction will use up more O2 to form NO and H2O, leading to a decrease in the partial pressure of O2.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in a reversible chemical reaction where the rates of the forward and reverse reactions are equal, resulting in no net change in the amounts of the reactants and products over time. It's essential to note that while the macroscopic properties, such as color and density, remain constant, the microscopic processes of the reactions continue; reactants are transformed into products and vice versa at equal rates. The equilibrium can be represented using a chemical equation with double arrows pointing in opposite directions.
Our example involves the gases NH3, O2, NO, and H2O. At equilibrium, 4 NH3 molecules react with 5 O2 molecules to form 4 NO molecules and 6 H2O molecules continuously, but the overall concentrations remain steady.A fundamental principle to remember is that equilibrium does not mean the reactants and products are in equal concentrations, but rather that their concentrations have stabilized and will not change unless the system is disturbed.
Partial Pressure
Partial pressure is a measure of the pressure exerted by a single type of gas in a mixture of gases. It's directly proportional to the concentration of that gas in the mixture. According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases.
For instance, in our chemical reaction equilibrium involving NH3, O2, NO, and H2O, each gas has its own partial pressure contributing to the total pressure of the gaseous system. If we alter the partial pressure of one component, for instance, NO, this would represent a change in concentration and therefore a disturbance to the equilibrium.
Equilibrium Shift
An equilibrium shift, also known as Le Chatelier's Principle, is an adjustment that a system at chemical equilibrium undergoes in response to an external stress, such as change in concentration, temperature, or pressure. The principle essentially states that a system will respond to such a stress by shifting in a direction that counteracts the change.
  • If the concentration of a product (like NO) is increased, the equilibrium will shift to the left (towards reactants) to decrease the concentration of NO, meaning more NH3 and O2 will be present. This is what we see in step 2 of the solution.
  • If the concentration of a reactant (like NH3) is decreased, the equilibrium will shift to the right (towards products) to increase the concentration of NH3 by producing more NO and H2O, consequently reducing the partial pressure of O2. This behavior is explained in step 3 of the solution.
Understanding these shifts is crucial for predicting how the equilibrium will respond to various changes, allowing chemists and students to manipulate and control reactions effectively.

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Most popular questions from this chapter

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 \(K_{e}\) ?

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

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{\circ}=+206 \mathrm{~kJ}\) (b) \(\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})=\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\), \(\Delta H^{2}=-41 \mathrm{~kJ}\) (c) \(2 \mathrm{SO}_{2}\) (g) \(+\mathrm{O}_{2}\) (g) \(=2 \mathrm{SO}_{3}\) (g), \(\Delta H^{\circ}=-198 \mathrm{~kJ}\)

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g), \Delta H^{\circ}=+57 \mathrm{~kJ}\) (b) \(\mathrm{X}_{2}(\mathrm{~g})=2 \mathrm{X}(\mathrm{g})\), where \(\mathrm{X}\) is a halogen (c) \(\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g})=\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g}), \Delta \mathrm{H}^{+}=-161 \mathrm{~kJ}\) (d) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H^{\circ}=-90 \mathrm{~kJ}\)

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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