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The formation of hydrogen sulfide from the elements is exothermic. $$ \mathrm{H}_{2}(\mathrm{~g})+\frac{1}{8} \mathrm{~S}_{8}(\mathrm{~s}) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-20.6 \mathrm{~kJ} / \mathrm{mol} $$ Predict the effect of each of these changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each change is made in a constant-volume system. (a) Adding more sulfur (b) Adding more \(\mathrm{H}_{2}\) (c) Raising the temperature

Short Answer

Expert verified
(a) Right, (b) Right, (c) Left.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts in the direction that counteracts the change. This principle helps predict how the equilibrium will adjust when different factors (like concentration, temperature, or pressure) are altered.
02

Analyze Adding More Sulfur

Adding more sulfur increases the concentration of a reactant (since sulfur is on the left side of the equation). According to Le Chatelier's Principle, the system will react to reduce this change by shifting the equilibrium to the right, favoring the production of more \( \mathrm{H}_{2} \mathrm{~S} \).
03

Analyze Adding More Hydrogen gas

Adding more \( \mathrm{H}_{2} \), another reactant, also increases reactant concentration. The equilibrium will shift to the right as the system attempts to reduce the increased concentration of \( \mathrm{H}_{2} \), leading to more formation of \( \mathrm{H}_{2} \mathrm{~S} \).
04

Effect of Raising the Temperature

The reaction is exothermic, as indicated by the negative enthalpy change (\( \Delta_{\mathrm{r}} H^{\circ} = -20.6 \mathrm{~kJ/mol} \)). Raising the temperature adds heat to the system. To reduce this added heat, the equilibrium will shift to the left, favoring the endothermic (reverse) reaction to absorb the excess heat.

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fascinating concept where a chemical reaction reaches a state where the concentrations of reactants and products remain constant over time. It's important to understand that this doesn't mean reactions stop all activity; they continuously occur, but at equal rates in both directions. This balanced state is dynamic, not static. For a reaction such as the formation of hydrogen sulfide, \[\mathrm{H}_{2}(\mathrm{g})+\frac{1}{8} \mathrm{~S}_{8}(\mathrm{s)} \rightleftharpoons \mathrm{H}_{2}\mathrm{~S}(\mathrm{g})\]equilibrium is achieved when the rate of the forward reaction (forming hydrogen sulfide) equals the rate of the backward reaction (breaking down hydrogen sulfide back to hydrogen and sulfur). Changes in conditions such as concentration, pressure, or temperature can shift this equilibrium position either to the left or right, altering the concentrations of reactants and products until a new equilibrium is established. It's like a delicate balance, where any change can tip the scale.
Exothermic Reaction
An exothermic reaction is one that releases energy, usually in the form of heat, to the surroundings. This occurs because the energy needed to form the product bonds is less than the energy released during the formation of reactant bonds. In simple terms, these reactions give off heat. For the reaction of forming hydrogen sulfide:- The enthalpy change, denoted as \( \Delta_{\mathrm{r}} H^{\circ} = -20.6 \mathrm{~kJ/mol} \), indicates the reaction releases 20.6 kJ per mole of heat.- Since it's exothermic, we think of heat as a product.Understanding exothermic reactions is crucial because the energy changes have practical implications for how a reaction is conducted and how equilibrium can be influenced by temperature. In day-to-day life, exothermic reactions are widespread, like burning fuels or even simple chemical hand warmers.
Effect of Temperature on Equilibrium
Temperature plays a significant role in affecting chemical equilibrium. According to Le Chatelier's Principle, if you change the temperature, the equilibrium position will shift in such a way as to counteract the change. For exothermic reactions:
  • Raising the temperature adds heat to the system.
  • The equilibrium shifts to the left, favoring the endothermic reverse reaction.
  • This shift occurs because the system tries to absorb the added heat.
In the context of hydrogen sulfide formation, increasing temperature will shift the equilibrium towards the reactants. This behavior is contrary to what might initially be desired if the goal is to produce more hydrogen sulfide. Temperature alterations thus provide a powerful tool to control the direction and yield of a chemical reaction, serving as a way to manipulate outcomes in industrial and laboratory settings.

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Most popular questions from this chapter

For each of these chemical reactions, predict whether the equilibrium constant at \(25^{\circ} \mathrm{C}\) is greater than 1 or less than \(1,\) or state that insufficient information is available. Also indicate whether each reaction is product-favored or reactant-favored. $$ \begin{array}{l} \text { (a) } 2 \mathrm{NaCl}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Na}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=823 \mathrm{~kJ} / \mathrm{mol} \\ \text { (b) } 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta_{\mathrm{r}} H^{\circ}=-566 \mathrm{~kJ} / \mathrm{mol} \\ \text { (c) } 3 \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \\ \Delta_{\mathrm{r}} H^{\circ}=2045 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Consider the gas-phase reaction of \(\mathrm{N}_{2}+\mathrm{O}_{2}\) to give \(2 \mathrm{NO}\) and the reverse reaction of 2 NO to give \(\mathbf{N}_{2}+\mathrm{O}_{2},\) discussed in Section 12-2e. An equilibrium mixture of \(\mathrm{NO}\), \(\mathrm{N}_{2}\), and \(\mathrm{O}_{2}\) at \(5000 . \mathrm{K}\) that contains equal concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) has a concentration of NO about half as great. Make qualitatively correct plots of the concentrations of reactants and products versus time for these two processes, showing the initial state and the final dynamic equilibrium state. Assume a temperature of \(5000 . \mathrm{K}\). Don't do any calculations-just sketch how you think the plots should look.

At elevated temperatures, \(\mathrm{BrF}_{5}\) establishes this equilibrium. $$ 2 \mathrm{BrF}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{Br}_{2}(\mathrm{~g})+5 \mathrm{~F}_{2}(\mathrm{~g}) $$ The equilibrium concentrations of the gases at \(1500 \mathrm{~K}\) are \(0.0064 \mathrm{~mol} / \mathrm{L}\) for \(\mathrm{BrF}_{5}, 0.0018 \mathrm{~mol} / \mathrm{L}\) for \(\mathrm{Br}_{2},\) and \(0.0090 \mathrm{~mol} / \mathrm{L}\) for \(\mathrm{F}_{2}\). Calculate \(K_{c}\)

A sealed 15.0 -L flask at \(300 . \mathrm{K}\) contains \(64.4 \mathrm{~g}\) of a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) in equilibrium. Calculate the total pressure in the flask. \(\left(\right.\) For \(2 \mathrm{NO}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) K_{\mathrm{P}}=\) 6.67 at \(300 . \mathrm{K} .)\)

The reaction of hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4},\) with chlorine trifluoride, \(\mathrm{ClF}_{3}\), has been used in experimental rocket motors. \(\mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+\frac{4}{3} \mathrm{ClF}_{3}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{HF}(\mathrm{g})+\mathrm{N}_{2}(\mathrm{~g})+\frac{2}{3} \mathrm{Cl}_{2}(\mathrm{~g})\) How is the equilibrium constant, \(K_{\mathrm{p}}\), for this reaction related to \(K_{\mathrm{p}}^{\prime}\) for the reaction written this way? $$ \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})+4 \mathrm{ClF}_{3}(\mathrm{~g}) \rightleftharpoons 12 \mathrm{HF}(\mathrm{g})+3 \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ (a) \(K_{\mathrm{P}}=K_{\mathrm{P}}^{\prime}\) (b) \(K_{\mathrm{P}}=1 / K_{\mathrm{P}}^{\prime}\) (c) \(K_{\mathrm{p}}^{3}=K_{\mathrm{P}}^{\prime}\) (d) \(K_{\mathrm{P}}=\left(K_{\mathrm{P}}^{\prime}\right)^{3}\) (e) \(3 K_{\mathrm{p}}=K_{\mathrm{P}}^{\prime}\)

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