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Consider the reaction $$\begin{array}{rl}{4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g)} & {\rightleftharpoons} \\ {4} & {\mathrm{NNO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{kJ}}\end{array}$$ Does each of the following increase, decrease, or leave unchanged the yield of \(\mathrm{NO}\) at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;(\mathbf{b})\) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\mathbf{c})\) decrease \(\left[\mathrm{O}_{2}\right] ;(\mathbf{d})\) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

Short Answer

Expert verified
(a) Increasing \(\mathrm{NH_3}\) concentration increases NO yield. (b) Increasing \(\mathrm{H_2O}\) concentration decreases NO yield. (c) Decreasing \(\mathrm{O_2}\) concentration decreases NO yield. (d) Decreasing container volume decreases NO yield. (e) Adding a catalyst leaves NO yield unchanged. (f) Increasing temperature decreases NO yield.

Step by step solution

01

(a) Increase the concentration of NH3

Increasing the concentration of NH3 will cause the system to shift towards the side that will reduce this change, according to Le Châtelier's principle. To counter this disturbance, the reaction will shift to the right, toward the products, resulting in an increased yield of NO. Conclusion: Increased yield of NO.
02

(b) Increase the concentration of H2O

Increasing the concentration of H2O will cause the system to shift towards the side that will reduce this change, according to Le Châtelier's principle. To counter this disturbance, the reaction will shift to the left, toward the reactants, resulting in a decreased yield of NO. Conclusion: Decreased yield of NO
03

(c) Decrease the concentration of O2

Decreasing the concentration of O2 will cause the system to shift towards the side that will counteract this change, according to Le Châtelier's principle. To counter this disturbance, the reaction will shift to the left, toward the reactants, resulting in a decreased yield of NO. Conclusion: Decreased yield of NO
04

(d) Decrease the volume of the container

Decreasing the volume of the container will increase the pressure of the system. According to Le Châtelier's principle, the reaction will shift to the side with fewer moles of gas to counteract the increased pressure. Since there are 9 moles of gas on the left (reactants) and 10 moles of gas on the right (products), the reaction will shift to the left, resulting in a decreased yield of NO. Conclusion: Decreased yield of NO
05

(e) Add a catalyst

Adding a catalyst will increase the rate of both the forward and reverse reactions equally without affecting the equilibrium position. Therefore, the yield of NO at equilibrium will remain unchanged. Conclusion: Unchanged yield of NO
06

(f) Increase the temperature

The reaction has a negative delta H, which means it is an exothermic reaction. According to Le Châtelier's principle, increasing the temperature will cause the reaction to shift to the side that will counteract this change, which in this case is the endothermic (reverse) direction. As a result, the reaction will shift to the left, resulting in a decreased yield of NO. Conclusion: Decreased yield of NO

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Key Concepts

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

Chemical Equilibrium
When a chemical reaction occurs, the reactants are converted into products, but under certain conditions, the products can also revert to reactants. Chemical equilibrium is the state in which the rate of the forward reaction equals the rate of the reverse reaction, meaning the concentrations of reactants and products remain constant over time.

This does not mean the amounts of reactants and products are equal, but that their concentrations no longer change. It's like a tug-of-war where both teams are equally strong; they pull with the same force, and the rope doesn't move. Understanding equilibrium is crucial because it tells us about a reaction's efficiency and how much product we can expect under given conditions.
Reaction Yield
The reaction yield refers to the amount of product formed in a chemical reaction. It is an essential factor in determining a chemical reaction's efficiency. In the context of our exercise, the amount of NO generated from the reaction involving NH3 and O2 is the reaction yield. The reaction yield of NO can change depending on various factors such as concentration of reactants, volume changes, addition of catalysts, or temperature shifts.

For example, increasing the concentration of NH3 directly impacts the yield of NO by driving the reaction towards more product formation. On the other hand, increasing the concentration of a product like H2O will drive the reaction back towards reactants, thus lowering the NO yield. Reaction yield is a vital concept as it influences industrial manufacturing processes, economic viability, and research directions in chemistry.
Equilibrium Shift
Le Châtelier's principle provides a prediction on how a system at equilibrium responds to changes in concentration, temperature, or pressure, leading to an 'equilibrium shift.' It essentially tells us that if a system at equilibrium is disturbed, it will adjust itself to minimize that disturbance.

Increasing reactants or decreasing products typically shifts equilibrium to the right, producing more products. Conversely, increasing products or decreasing reactants shifts equilibrium to the left, forming more reactants. Changes in volume and pressure also affect the equilibrium position; reducing volume (which increases pressure) shifts the balance toward the side with fewer moles of gas. Temperature changes can shift equilibrium too; for exothermic reactions, increasing temperature shifts equilibrium to the left, decreasing product yield. These shifts are at the heart of understanding how to control and manipulate chemical reactions for desired outcomes.

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Most popular questions from this chapter

The equilibrium \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) is established at 500 \(\mathrm{K}\) . An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for \(\mathrm{NO}, \mathrm{Cl}_{2},\) and \(\mathrm{NOCl}\) , respectively. (a) Calculate \(K_{p}\) for this reaction at 500.0 \(\mathrm{K}\) . (b) If the vessel has a volume of 5.00 \(\mathrm{L}\) , calculate \(K_{c}\) at this temperature.

At a temperature of 700 \(\mathrm{K}\) , the forward and reverse rate constants for the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\) are \(k_{f}=1.8 \times 10^{-3} \mathrm{M}^{-1} \mathrm{s}^{-1}\) and \(k_{r}=0.063 \mathrm{M}^{-1} \mathrm{s}^{-1}\) (a) What is the value of the equilibrium constant \(K_{c}\) at 700 \(\mathrm{K} ?\) (b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of \(k_{f}=0.097 M^{-1} \mathrm{s}^{-1}\) and \(k_{r}=2.6 \mathrm{M}^{-1} \mathrm{s}^{-1}\) at 800 \(\mathrm{K} ?\)

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ An equilibrium mixture in a 5.00 -L vessel at \(100^{\circ} \mathrm{C} \) contains 3.22 \(\mathrm{g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{g}\) of \(\mathrm{NO},\) and 4.19 \(\mathrm{g}\) of Br \(_{2}\) .(a) Calculate \(K_{c}\) (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of \(\mathrm{NOBr}\) ?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .\) An equilibrium mixture in a 2.00 -L vessel is found to contain 0.0406 \(\mathrm{mol}\) \(\mathrm{CH}_{3} \mathrm{OH}, 0.170 \mathrm{mol} \mathrm{CO},\) and 0.302 \(\mathrm{mol} \mathrm{H}_{2}\) at 500 \(\mathrm{K}\) . Calculate \(K_{c}\) at this temperature.

The following equilibria were measured at 823 K: \begin{equation} \begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{aligned} \end{equation} (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\operatorname{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at 823 \(\mathrm{K}\) . (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{K} ?\) (c) If you were to place 5.00 \(\mathrm{g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of 250 \(\mathrm{mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of 1.00 atm and a temperature of \(298 \mathrm{K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to 823 \(\mathrm{K}\) and allowed to come to equilibrium, how much \(\mathrm{CoO}(s)\) remains?

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