A mixture consisting of \(2.23 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2}\) and \(6.69 \times 10^{-3} \mathrm{~mol} \mathrm{H}_{2}\) in a \(500-\mathrm{ml}\). container was heated to \(600 \mathrm{~K}\) and allowed to reach equilibrium. Will more ammonia be formed if that equilibrium mixture is then heated to \(700 \mathrm{~K}\) ? For \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})=\) \(2 \mathrm{NH}_{3}(\mathrm{~g}), K=1.7 \times 10^{-3}\) at \(600 \mathrm{~K}\) and \(7.8 \times 10^{-5}\) at \(700 \mathrm{~K}\).

Short Answer

Expert verified
No, more ammonia will not be formed; instead, the amount of ammonia will decrease when the equilibrium mixture is heated to 700K, because the reaction is exothermic.

Step by step solution

01

Identify the Reaction and its Equilibrium Constant

The chemical reaction given is \(N_{2(g)} + 3H_{2(g)} = 2NH_{3(g)}\). The equilibrium constant (K) changes with temperature; K is \(1.7 \times 10^{-3}\) at 600K and \(7.8 \times 10^{-5}\) at 700K. A decrease in K with an increase in temperature suggests the reaction is exothermic.
02

Determine the Direction of the Equilibrium Shift

According to Le Châtelier's Principle, increasing the temperature of an exothermic reaction will shift the equilibrium to favor the reactants. Consequently, heating the equilibrium mixture from 600K to 700K will result in less ammonia, \(NH_{3}\), and more of the reactants, \(N_{2}\) and \(H_{2}\).
03

Conclude the Effect of Temperature on Ammonia Formation

As the reaction favors the formation of reactants at higher temperatures for an exothermic reaction, and since the equilibrium constant at 700K is lower than at 600K, it is concluded that heating the mixture to 700K will result in less ammonia being formed at equilibrium.

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Key Concepts

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

Le Châtelier's Principle
Le Châtelier's Principle offers insight into what happens to a chemical system when it experiences a change in concentration, pressure, volume, or temperature. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system adjusts itself to partially counteract the effect of the disturbance and a new equilibrium is established.

For instance, consider a sealed container holding a mixture of gases at equilibrium. When the temperature increases, the system will favor the reaction that absorbs heat, essentially trying to reduce the impact of the temperature rise. Likewise, reducing the temperature causes the system to shift toward the reaction that releases heat, which increases the temperature inside the system. This principle helps predict how concentration and pressure affect equilibria as well, critical for understanding reactions and optimizing industrial processes.
Equilibrium Constant
The equilibrium constant, represented by the symbol K, is a number that provides a measure of the extent of a reaction at equilibrium at a given temperature. It is determined by the concentrations of the products raised to their stoichiometric coefficients divided by the concentrations of the reactants raised to their stoichiometric coefficients.

In mathematical terms, for a reaction like \(aA + bB \leftrightarrow cC + dD\), the equilibrium constant \(K\) is expressed as \[K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\] where \[\text{[A],[B],[C],[D]}\] are the concentrations of the respective chemicals at equilibrium. It's important to note that K is dimensionless and its value is constant only at a constant temperature; it changes when the temperature changes.
Effect of Temperature on Equilibrium
Temperature has a profound impact on chemical equilibrium. For exothermic reactions, which release heat, an increase in temperature causes the equilibrium position to shift towards the reactants, as predicted by Le Châtelier's principle. Conversely, for endothermic reactions, which absorb heat, raising the temperature drives the equilibrium towards the products. This change is reflected in the value of the equilibrium constant. If the constant decreases with an increase in temperature, this indicates an exothermic reaction, suggesting that heat is a product. On the other hand, if K increases as the temperature rises, the reaction is likely endothermic, treating heat as a reactant.

Understanding how temperature affects chemical equilibrium is essential for chemical engineering and industrial applications, where precise control of reaction conditions can substantially impact the efficiency and yield of chemical processes.
Exothermic Reaction
An exothermic reaction is a chemical process that releases energy in the form of heat or light. The classic illustration of such a reaction is the combustion of fuels, which produces heat and sometimes light. Another everyday example is the rusting of iron, which releases heat slowly and may not be as perceptible as combustion.

In terms of thermodynamics, the energy released in exothermic reactions is typically because the energy required to break the bonds in the reactants is less than the energy released when new bonds form in the products. In an energy diagram, exothermic reactions show the products as having lower energy than the reactants, with the difference corresponding to the energy released to the surroundings. These reactions often occur spontaneously and are utilized in various applications, from heating homes to powering vehicles.

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Most popular questions from this chapter

At \(2500 \mathrm{~K}, K_{e}=20\) for the reaction \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{ClF}(\mathrm{g}) .\) An analysis of a reaction vessel at \(2500 \mathrm{~K}\) revealed the presence of \(0.18 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) \(\mathrm{Cl}_{2}, 0.31 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{~F}_{2}\), and \(0.92 \mathrm{~mol} \cdot \mathrm{L}^{-1} \mathrm{ClF}\). Will \(\mathrm{ClF}\) tend to form or to decompose as the reaction proceeds toward cquilibrium?

Write the reaction quotients \(Q_{c}\) for (a) \(\mathrm{Cu}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CuCl}_{2}(\mathrm{~s})\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{2}\) (s) \(\rightarrow \mathrm{N}_{2} \mathrm{O}\) (g) \(+2 \mathrm{H}_{2} \mathrm{O}\) (g) (c) \(\mathrm{MgCO}_{3}(\mathrm{~s}) \rightarrow \mathrm{MgO}\) (s) \(+\mathrm{CO}_{2}\) (g)

If \(Q=1.0\) for the reaction \(N_{2}(g)+O_{2}(g) \rightarrow\) \(2 \mathrm{NO}(\mathrm{g})\) at \(25^{\circ} \mathrm{C}\), will the reaction have a tendency to form products or reactants, or will it be at equilibrium?

A \(0.10-\mathrm{mol}\) sample of pure czone, \(\mathrm{O}_{3}\), is placed in a sealed \(1.0\) - L container and the reaction \(2 \mathrm{O}_{3}(\mathrm{~g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{~g})\) is allowed to reach equilihrium. A \(0.50-\mathrm{mol}\) sample of pure ozone is placed in a second 1.0-L container at the same temperature and allowed to reach equilibrium. Without doing any calculations, predict which of the following will be different in the two containers at equilibrium. Which will be the same? Explain each of your answers: (a) amount of \(\mathrm{O}_{2}\); (b) concentration of \(\mathrm{O}_{2}\); (c) the ratio \(\left[\mathrm{O}_{2}\right] /\left[\mathrm{O}_{3}\right]\); (d) the ratio \(\left.\left[\mathrm{O}_{2}\right]^{3} / \mathrm{O}_{3}\right]^{2} ;\) (e) the ratio \(\left|\mathrm{O}_{3}\right|^{2} /\left[\mathrm{O}_{2}\right]^{3}\).

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) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H^{\circ}=-90 \mathrm{~kJ}\)

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