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Consider the system \(4 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-1530.4 \mathrm{~kJ}\) (a) How will the amount of ammonia at equilibrium be affected by (1) removing \(\mathrm{O}_{2}(g)\) ? (2) adding \(\mathrm{N}_{2}(g)\) ? (3) adding water? (4) expanding the container at constant pressure? (5) increasing the temperature? (b) Which of the above factors will increase the value of \(K ?\) Which will decrease it?

Short Answer

Expert verified
Answer: The factors and their effects on the amount of ammonia at equilibrium are as follows: 1. Removing O2 will increase the amount of ammonia at equilibrium. 2. Adding N2 will also increase the amount of ammonia at equilibrium. 3. Adding water will not have a significant effect on the amount of ammonia at equilibrium. 4. Expanding the container at constant pressure will cause an increase in the amount of ammonia at equilibrium. 5. Increasing the temperature will increase the amount of ammonia at equilibrium. The equilibrium constant K is affected only when the temperature is increased, causing it to decrease for this exothermic reaction. The other factors do not affect the value of K, although they can cause the equilibrium to shift.

Step by step solution

01

Removing O2

According to Le Chatelier's principle, removing O2 would cause the equilibrium to shift towards the reactants to replace the lost O2. Therefore, the amount of ammonia at equilibrium will increase.
02

Adding N2

According to Le Chatelier's principle, adding N2 would cause the equilibrium to shift away from the products in order to lower the concentration of N2 in the system. As a result, the amount of ammonia at equilibrium will also increase.
03

Adding water

Adding water will not have a significant effect on the amount of ammonia in the equilibrium because water is in the liquid phase, and the reaction occurs in the gas phase. The concentrations of gases are not affected by the presence of liquids.
04

Expanding the container at constant pressure

Expanding the container at constant pressure will cause the equilibrium to shift towards the side with a higher number of moles of gas. In this reaction, there are a total of 4 moles of gas on the reactant side and 2 moles of gas on the product side. Therefore, expanding the container will cause the equilibrium to shift towards the reactants, increasing the amount of ammonia at equilibrium.
05

Increasing the temperature

Increasing the temperature for an exothermic reaction will cause the equilibrium to shift towards the reactants to absorb the added heat, as predicted by Le Chatelier's principle. Consequently, the amount of ammonia at equilibrium will increase. (b)
06

Effects on equilibrium constant K

1. Removing O2 does not affect K. The equilibrium shifts, but the value of K remains constant. 2. Adding N2 also does not affect K. The equilibrium shifts, but the value of K remains constant. 3. Adding water does not affect K since it does not change the concentrations of gases in the system. 4. Expanding the container at constant pressure does not affect K. 5. Increasing the temperature will decrease the value of K for an exothermic reaction. This is because, according to the Van't Hoff equation, the equilibrium constant is inversely proportional to temperature for an exothermic process. In conclusion, only increasing the temperature affects the equilibrium constant K, causing it to decrease for this particular exothermic reaction. The other factors will cause the equilibrium to shift, but they do not affect the value of K.

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Key Concepts

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

Chemical Equilibrium
Understanding chemical equilibrium is essential for anyone studying reactions. Imagine a seesaw perfectly balanced with two children of equal weight. Now, replace the children with chemical reactants and products, and you've got a snapshot of chemical equilibrium. At this point, the rate of the forward reaction (reactants turning into products) equals the rate of the reverse reaction (products turning back into reactants). This doesn't mean the amounts of reactants and products are equal, but that they're stable over time, creating a dynamic balance.

Example in Action

The equilibrium we're examining involves the substances NH3, O2, N2, and H2O. Changes to the system, such as removing or adding a substance, will prompt the equilibrium to shift, aiming to counteract those changes and regain balance—much like how a child shifting their weight on a seesaw causes the other side to move in response.
Equilibrium Constant (K)
The equilibrium constant, denoted as K, is the heart of chemical equilibrium. It quantifies the balance between reactants and products in a reversible reaction at a given temperature. Calculated using the concentrations of the entities involved, the constant helps chemists predict the direction and extent of chemical reactions.

Significance of K

A higher K value signifies a greater concentration of products compared to reactants at equilibrium, hinting that the reaction favors product formation. Conversely, a lower K suggests the reaction leans towards the reactants. It's important to realize that K doesn't change just because the amounts of reactants or products change. It's temperature that K is sensitive to, so alterations in heat can lead to shifts in K's value and consequently signal a difference in the favored direction of the reaction.
Exothermic Reactions
Exothermic reactions are like cozy fireplaces, releasing warmth into their surroundings. During these reactions, energy is given off, often in the form of heat, making the total energy of the products less than the reactants. Products in exothermic reactions are typically more stable than the reactants because they have released energy.

Visualizing Exothermicity

One classic example is the combustion of natural gas in a furnace. Once initiated, it releases heat that warms your home. In our textbook reaction, when NH3 and O2 reacts to form N2 and H2O, it is also exothermic, and this characteristic plays a major role in how temperature changes affect the reaction's equilibrium.
Effects of Temperature on Equilibrium
Temperature can be the wizard of the equilibrium realm, casting spells that change the balance of a chemical reaction. Think of temperature as the energy currency of molecules; a rise in temperature means there's more energy available for reactions. In an exothermic reaction, adding heat is akin to adding more products, because heat is one of the 'products' released. This nudges the equilibrium to favor the reactants to absorb the excess energy.

For the reaction with NH3 and O2, if we increase the temperature, Le Chatelier's principle informs us that the system counteracts the added heat by favoring the reverse reaction. Resulting in more NH3 and less N2 and H2O being formed. As a result, the reaction will shift to the left, and in this case, simultaneously decrease the equilibrium constant, K, highlighting the profound effect temperature has on chemical equilibrium.

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Most popular questions from this chapter

When carbon monoxide reacts with hydrogen gas, methane and steam are formed. $$\mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ At \(1127^{\circ} \mathrm{C}\), analysis at equilibrium shows that \(P_{\mathrm{CO}}=0.921 \mathrm{~atm}\), \(P_{\mathrm{H}_{2}}=1.21 \mathrm{~atm}, P_{\mathrm{CH}_{4}}=0.0391 \mathrm{~atm}\), and \(P_{\mathrm{H}_{2} \mathrm{O}}=0.0124 \mathrm{~atm} .\) What is the equilibrium constant, \(K\), for the reaction at \(1127^{\circ} \mathrm{C}\) ?

At \(25^{\circ} \mathrm{C}, K=2.2 \times 10^{-3}\) for the reaction $$\mathrm{ICl}(g) \rightleftharpoons \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g)$$ Calculate \(K\) at \(25^{\circ} \mathrm{C}\) for (a) the decomposition of ICl into one mole of iodine and chlorine. (b) the formation of two moles of \(\operatorname{ICl}(g)\).

Consider the system $$\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta H=98.9 \mathrm{~kJ}$$ (a) Predict whether the forward or reverse reaction will occur when the equilibrium is disturbed by (1) adding oxygen gas. (2) compressing the system at constant temperature. (3) adding argon gas. (4) removing \(\mathrm{SO}_{2}(g)\). (5) decreasing the temperature. (b) Which of the above factors will increase the value of \(K ?\) Which will decrease it?

At a certain temperature, \(K\) is \(1.3 \times 10^{5}\) for the reaction $$2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g)$$ What is the equilibrium pressure of hydrogen sulfide if those of hydrogen and sulfur gases are \(0.103\) atm and \(0.417\) atm, respectively?

The following data are for the system $$\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)$$ $$\begin{array}{ccccccc}\hline \text { Time (s) } & 0 & 20 & 40 & 60 & 80 & 100 \\ P_{\mathrm{A}} \text { (atm) } & 1.00 & 0.83 & 0.72 & 0.65 & 0.62 & 0.62 \\ P_{\mathrm{B}} \text { (atm) } & 0.00 & 0.34 & 0.56 & 0.70 & 0.76 & 0.76 \\ \hline\end{array}$$ Prepare a graph of \(P_{\Lambda}\) and \(P_{\mathrm{B}}\) versus time and use it to answer the following questions: (a) Estimate \(P_{\mathrm{A}}\) and \(P_{\mathrm{g}}\) after \(30 \mathrm{~s}\). (b) Estimate \(P_{\mathrm{A}}\) after \(150 \mathrm{~s}\). (c) Estimate \(P_{\mathrm{B}}\) when \(P_{\mathrm{A}}=0.700 \mathrm{~atm}\).

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