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Explain how each of the following affects the amount of \(\mathrm{H}_{2}\) present in an equilibrium mixture in the reaction \(3 \mathrm{Fe}(\mathrm{s})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}_{3} \mathrm{O}_{4}(\mathrm{s})+4 \mathrm{H}_{2}(\mathrm{g})\) $$ \Delta H^{\circ}=-150 \mathrm{kJ} $$ (a) Raising the temperature of the mixture; (b) introducing more \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) ;\) (c) doubling the volume of the container holding the mixture; (d) adding an appropriate catalyst.

Short Answer

Expert verified
(a) Raising the temperature of the mixture will decrease the amount of \(H_2\). (b) Introducing more \(H_2O\) will increase the amount of \(H_2\). (c) Doubling the volume of the container holding the mixture will also increase the amount of \(H_2\). (d) Adding an appropriate catalyst will not change the amount of \(H_2\) at equilibrium.

Step by step solution

01

Understanding the Effect of Temperature

According to Le Chatelier's Principle, an increase in temperature shifts the equilibrium towards the endothermic side of a reaction, in order to absorb the extra heat. Given that the reaction \(\Delta H^{\circ}=-150 \mathrm{kJ}\) is exothermic (heat is released), raising the temperature of the mixture will move the equilibrium to the left, thus decreasing the amount of hydrogen (\(H_2\)).
02

Understanding the Effect of Introducing More \(H_2O\)

Adding more of a reactant to a system in equilibrium will shift the equilibrium towards the products, to consume the added substance. Therefore, introducing more \(H_2O\) will shift the equilibrium to the right, which will increase the amount of hydrogen (\(H_2\)).
03

Understanding the Effect of Doubling the Volume

Changing the volume affects the pressure of a system in equilibrium. By increasing the volume (or decreasing the pressure), the system will shift towards the side with a greater number of gaseous molecules to counteract the change. In this case, the right side of the equation has more gas molecules (4 molecules of \(H_2\)) than the left side. So doubling the volume will shift the equilibrium to the right and thus increase the amount of hydrogen (\(H_2\)).
04

Understanding the Effect of Adding a Catalyst

Adding a catalyst to a reaction does not change the position of the equilibrium; it simply speeds up the rate at which equilibrium is reached. Therefore, adding a catalyst will not affect the amount of hydrogen (\(H_2\)) at equilibrium.

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle, named after the French chemist Henri Louis Le Chatelier, is a key concept in chemical equilibrium that helps predict how a system at equilibrium reacts to external changes. It states that if a dynamic equilibrium is disturbed by changing the conditions, such as concentration, temperature, or pressure, the system adjusts itself to counteract the change and a new equilibrium is established.

For example, if the concentration of a reactant in a reaction mixture is increased, according to Le Chatelier's Principle, the system shifts its equilibrium position to consume the excess reactants and produce more products, thereby reducing the impact of the disturbance. This adjustment continues until a new equilibrium position is reached where the rates of the forward and backward reactions are equal again.
Temperature Effects on Equilibrium
When the temperature of a system at equilibrium changes, it effects the balance between the exothermic (releases heat) and endothermic (absorbs heat) reactions. For an exothermic reaction, increasing the temperature shifts the equilibrium position to favor the formation of reactants (leftward shift), as the system attempts to absorb the added heat. Conversely, in an endothermic reaction, an increase in temperature causes the system to shift towards the products (rightward shift).

More technically, this is because temperature changes alter the reaction quotient, K, which compares the concentrations of the reactants and products at a given moment. Raising temperature increases K for an endothermic reaction and decreases it for an exothermic reaction, leading to the shifts in the equilibrium.
Equilibrium and Pressure Changes
Pressure changes can have a significant impact on the position of equilibrium for reactions involving gases. According to Le Chatelier's Principle, increasing the pressure of a system by decreasing its volume causes the equilibrium position to shift towards the side with fewer gas molecules. This is because the system aims to reduce the pressure increase by favoring the formation of less gaseous particles.

On the other hand, if the volume is increased and the pressure is decreased, the equilibrium will favor the side with more gas molecules, trying to increase the pressure again. This is a crucial concept for processes that involve the synthesis of gaseous products, where manipulating pressure can increase yield.
Catalysts and Equilibrium
A catalyst is a substance that speeds up the rate of a chemical reaction without being consumed in the process. It works by providing an alternative reaction pathway with a lower activation energy.

Importantly, while a catalyst can help a reaction reach equilibrium more quickly, it does not affect the position of the equilibrium itself. Regardless of the presence of a catalyst, the relative concentrations of reactants and products at equilibrium remain the same. This means that while a catalyst is incredibly useful in industry and research for speeding up reactions and reducing energy costs, it cannot be used to alter the equilibrium position to favor the production of more products or reactants.

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Most popular questions from this chapter

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

In the equilibrium described in Example \(15-12,\) the percent dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be expressed as $$\frac{3.00 \times 10^{-3} \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}}{0.0240 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4} \text { initially }} \times 100 \%=12.5 \%$$ What must be the total pressure of the gaseous mixture if \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})\) is to be \(10.0 \%\) dissociated at \(298 \mathrm{K} ?\) $$ \mathrm{N}_{2} \mathrm{O}_{4} \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.113 \text { at } 298 \mathrm{K} $$

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.

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