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Consider \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) \(\Delta H=-904.4 \mathrm{~kJ}\). How does each of the following changes affect the yield of \(\mathrm{NO}\) at equilibriunt? Answer increase, decrease, or no change: (a) increase [NII \(_{3}\) ]; (b) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;\) (c) decrease \(\left[\mathrm{O}_{2}\right.\) \\} (d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.

Short Answer

Expert verified
(a) Increase [NH3]: The yield of NO will increase. (b) Increase [H2O]: The yield of NO will decrease. (c) Decrease [O2]: The yield of NO will decrease. (d) Decrease the volume: The yield of NO will decrease. (e) Add a catalyst: No change in the yield of NO. (f) Increase temperature: The yield of NO will decrease.

Step by step solution

01

Reaction Assessment

Before solving each part, we must identify the number of moles of reactants and products in the reaction. Reactants: 4 moles of NH3 and 5 moles of O2. Products: 4 moles of NO and 6 moles of H2O.
02

a) Increase [NH3]

According to Le Chatelier's Principle, if the concentration of a reactant is increased, the reaction will shift to the side that consumes the added reactant. In this case, increasing NH3 will shift the reaction to the product side. So, the yield of NO will increase.
03

b) Increase [H2O]

If the concentration of a product is increased, the reaction will shift to the side that consumes the added product. In this case, increasing H2O will shift the reaction to the reactant side. So, the yield of NO will decrease.
04

c) Decrease [O2]

If the concentration of a reactant is decreased, the reaction system will shift to the side that produces the removed reactant. In this case, decreasing O2 (reactant) will shift the reaction to the reactant side. So, the yield of NO will decrease.
05

d) Decrease the volume of the container

When the volume of the container is decreased, pressure increases. In this case, the reaction will shift to the side with fewer moles of gas to minimize pressure increase. Since 4 + 6 (product side) > 4 + 5 (reactant side), the reaction will shift to the reactant side, and the yield of NO will decrease.
06

e) Add a catalyst

A catalyst only increases the rate of the reaction and helps the system reach equilibrium faster. It doesn't affect the position of the equilibrium or the yield of the products. So, the yield of NO will have no change.
07

f) Increase temperature

Since the reaction enthalpy ΔH is negative (-904.4 kJ), it indicates that the reaction is exothermic. According to Le Chatelier's Principle, increasing the temperature in an exothermic reaction will cause the system to counteract the temperature increase by shifting to the endothermic direction. In this case, the endothermic direction is the reactant side (backwards). Therefore, the yield of NO will decrease.

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Key Concepts

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

Equilibrium Shifts
Le Chatelier's Principle helps us understand how changes in a system affect chemical equilibrium. When we make changes to a system, like tweaking concentrations or pressure, the equilibrium shifts to counteract these changes and restore balance.

Here's a simplified way of thinking about it:
  • Increase the concentration of a reactant? The equilibrium shifts to form more products.
  • Increase the concentration of a product? The equilibrium shifts to form more reactants.
  • Decrease the concentration of a reactant? The equilibrium shifts to form more reactants.
  • Decrease the concentration of a product? The equilibrium shifts to make more products.
By understanding these shifts, we can predict and manipulate how much of a product, like NO, we get from a reaction.
Reaction Yield
In any chemical reaction, the yield refers to the amount of product formed. Maximizing yield is often a key goal in industrial chemical processes.

What factors can influence the yield?
  • Reactant Concentration: Increasing the concentration of reactants pushes the reaction towards more product formation.
  • Product Concentration: Increasing product concentration can decrease the yield because the equilibrium tries to return to balance by forming more reactants.
  • Volume and Pressure: A decrease in volume increases pressure, which affects the position of equilibrium depending on the number of moles of gases involved.
Understanding these variables can help control the amount of product like NO produced at equilibrium.
Exothermic Reaction
Exothermic reactions release heat as they proceed, which can be a crucial factor affecting equilibrium. The given reaction has a negative equation the reaction releases energy. When you increase the temperature of an exothermic reaction, the system tries to cool itself by favoring the reverse reaction (forming reactants).

Here's how it works:
  • In an exothermic reaction, if you increase the temperature, the yield of the product (like NO) typically decreases because the equilibrium shifts towards the reactant side.
  • Conversely, decreasing the temperature will increase the yield, pushing the equilibrium towards the product side.
This temperature dependence is pivotal in controlling reactions, especially when optimizing the desired product yield.
Catalysis Effects
Catalysts are substances that speed up the rate of a chemical reaction without being consumed. They are like facilitators, making it easier for the conversion from reactants to products and vice versa.

How do catalysts work?
  • They lower the activation energy required for a reaction, letting the process happen faster.
  • Catalysts do not alter the position of equilibrium; they only help the system reach equilibrium more quickly.
For example, in the reaction forming NO, adding a catalyst won't change the yield of NO. It will just ensure that the reaction reaches equilibrium faster, which can be advantageous in industrial processes to save time and energy.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$ \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use thermochemical data in Appendix \(C\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

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 mix- ture 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 \(500 \mathrm{~K}\). Cal- culate \(K_{c}\) at this temperature.

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00-L flask at \(303 \mathrm{~K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Calculate \(K_{c}\) for this reaction at this temperature.

(a) At \(800 \mathrm{~K}\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0-L vessel contains \(2.67 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{I}(\mathrm{g})\), how many grams of \(I_{2}\) are in the mixture? (b) For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \quad K_{p}=3.0 \times 10^{4}\) \(700 \mathrm{~K}\). In a 2.00-L vessel the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

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