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The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ is an endothermic process. Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a flask in which this equilibrium exists, how is the equilibrium affected? What if some additional \(\mathrm{NH}_{3}\) is placed in the flask? What will happen to the pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{S}\) is removed from the flask?

Short Answer

Expert verified
Increasing temperature shifts equilibrium right. Adding \(\mathrm{NH}_{4} \mathrm{HS}\) or removing \(\mathrm{H}_{2} \mathrm{S}\) shifts equilibrium right; adding \(\mathrm{NH}_{3}\) shifts it left.

Step by step solution

01

Understand Le Chatelier's Principle

Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, volume, or pressure, the equilibrium will shift in such a way as to counteract that change and restore a new equilibrium.
02

Analyze the Effect of Temperature Increase

Since the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) into \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) is endothermic, increasing the temperature provides more energy to the system. According to Le Chatelier's principle, the equilibrium will shift to the right (towards products) to absorb the additional heat.
03

Adding More NH4HS

Adding more \(\mathrm{NH}_{4} \mathrm{HS}\) increases the concentration of the reactant. The equilibrium will shift to the right to decrease the concentration of \(\mathrm{NH}_{4} \mathrm{HS}\) by converting it into more \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\).
04

Adding More NH3

Adding \(\mathrm{NH}_{3}\) increases the concentration of a product. To counteract the increase, the equilibrium will shift to the left, towards forming more \(\mathrm{NH}_{4} \mathrm{HS}\) and using up some of the added \(\mathrm{NH}_{3}\).
05

Removing H2S

Removing \(\mathrm{H}_{2} \mathrm{S}\) decreases the concentration of a product. As a result, the equilibrium will shift to the right, producing more \(\mathrm{H}_{2} \mathrm{S}\) and \(\mathrm{NH}_{3}\) to restore balance, thus increasing the pressure of \(\mathrm{NH}_{3}\).

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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 state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of the reactants and products remain constant over time, not because the reactions have stopped, but because they are occurring at the same rate. It is a dynamic equilibrium, meaning that molecules are constantly reacting, but overall quantities remain stable.
When a chemical system reaches equilibrium, it does not imply that the reactants and products are in equal concentrations. Instead, equilibrium is reached when their rates stabilize. Le Chatelier's Principle is a useful tool to understand how the equilibrium position can change. When external changes, such as temperature or concentration shifts, are applied to the system, the equilibrium "tries" to counteract the effect of the change. Let us delve deeper into how these changes particularly affect endothermic reactions.
Endothermic Reaction
An endothermic reaction is a chemical reaction that absorbs heat from its surroundings. These reactions cool the surrounding area as they require energy to proceed. In the context of the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) into \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\), this process is endothermic, meaning it absorbs heat.
When we apply Le Chatelier's Principle, if we increase the temperature in an endothermic reaction, the equilibrium will shift towards the products.
This is because the system absorbs the added heat to minimize the temperature change, favoring the formation of more products. Therefore, raising the temperature effectively "pushes" the equilibrium to the right in the equation, helping to counterbalance the increased temperature.
The understanding of endothermic reactions and their behavior under varying temperature conditions helps predict reaction direction and product formation.
Effect of Concentration Changes
The concentration of reactants or products in a chemical reaction can have a profound effect on the position of equilibrium. Imagine equilibrium as a seesaw, where changes in concentration on one side will shift the balance of the system in order to maintain equilibrium.
Let's take the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\):
  • Adding more \(\mathrm{NH}_{4} \mathrm{HS}\) increases the concentration of reactants. This causes the equilibrium to shift towards the right, producing more \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) to restore balance.
  • Introducing more \(\mathrm{NH}_{3}\) means increasing the concentration of one of the products, resulting in a shift in equilibrium towards the left. This helps "use up" the added \(\mathrm{NH}_{3}\) by forming more \(\mathrm{NH}_{4} \mathrm{HS}\).
  • Removing some \(\mathrm{H}_{2} \mathrm{S}\) decreases the concentration of a product, pushing the equilibrium towards the right. This attempt to balance the concentration allows for more \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) generation.
Comprehending these changes helps predict how various concentrations can influence reaction dynamics and equilibrium positioning.
Temperature Effects on Equilibrium
Temperature changes are a crucial factor in influencing chemical equilibria, often altering the direction of equilibrium shifts. For reactions like the endothermic decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\), temperature plays a key role.
When increasing the temperature, the equilibrium of an endothermic reaction shifts towards the products.
This shift occurs because absorbing additional heat helps the reaction proceed in the direction that uses up this energy, similar to a sponge soaking up water. Conversely, if the temperature is decreased, the activity of the reaction reverses. The equilibrium is driven towards the reactants to release heat into the system, thus minimizing the temperature reduction. This balance is a prime example of how Le Chatelier's Principle predicts equilibrium adjustments in response to temperature variations. By studying the effects of temperature, one can effectively determine the control needed to drive the reaction towards desired products or reactants.

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Most popular questions from this chapter

Calculate \(K\) for the reaction $$ \mathrm{SnO}_{2}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) $$ given the following information: $$\begin{array}{c} \mathrm{SnO}_{2}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K=8.12 \\\ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \quad K=0.771 \end{array}$$

At 2300 K the equilibrium constant for the formation of \(\mathrm{NO}(\mathrm{g})\) is \(1.7 \times 10^{-3}\) $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ (a) Analysis shows that the concentrations of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are both \(0.25 \mathrm{M},\) and that of \(\mathrm{NO}\) is \(0.0042 \mathrm{M}\) under certain conditions. Is the system at equilibrium? (b) If the system is not at equilibrium, in which direction does the reaction proceed? (c) When the system is at equilibrium, what are the equilibrium concentrations?

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at a high temperature contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=3.77 \times 10^{-3} \mathrm{mol} / \mathrm{L}\) \(\left[\mathrm{O}_{2}\right]=4.30 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{SO}_{3}\right]=4.13 \times\) \(10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate the equilibrium constant, \(K_{c}\) for the reaction. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

The equilibrium constant \(K\) for the reaction $$ \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is \(6.66 \times 10^{-12}\) at 1000 K. Calculate \(K\) for the reaction $$ 2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}_{2}(\mathrm{g}) $$

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left|\mathrm{SO}_{2}\right|=\) $$5.0 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left|\mathrm{O}_{2}\right|=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L}, \text { and } $$ \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium?$$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) K_{c}=279 $$

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