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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 to products; adding \( \mathrm{NH}_{4} \mathrm{HS} \) has little effect; adding \( \mathrm{NH}_3 \) shifts to reactants; removing \( \mathrm{H}_{2} \mathrm{S} \) increases \( \mathrm{NH}_{3} \) pressure.

Step by step solution

01

Understanding the reaction and Le Chatelier's principle

The decomposition of \( \mathrm{NH}_{4} \mathrm{HS} \) is an endothermic reaction, meaning that it absorbs heat. According to Le Chatelier's principle, if a change is imposed on a system at equilibrium, the system will shift to counteract this change. For an endothermic reaction like this, increasing the temperature adds heat, causing the equilibrium to shift toward the products \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \).
02

Effects of Adding More NH4HS

Adding more \( \mathrm{NH}_{4} \mathrm{HS} \) to the mixture provides more reactant. Since \( \mathrm{NH}_{4} \mathrm{HS} \) is an equilibrium solid, its concentration does not affect the position of equilibrium under usual conditions. Thus, adding more will not shift the equilibrium; however, more products can form if the reaction proceeds, keeping the equilibrium constant \( K_e \) unchanged.
03

Effects of Introducing Additional NH3

Adding more \( \mathrm{NH}_3 \) increases the concentration of one of the products. According to Le Chatelier's principle, the equilibrium will shift toward the reactants to counteract this increase in product concentration. Thus, more \( \mathrm{NH}_{4} \mathrm{HS} \) will form, lowering \( \mathrm{NH}_3 \)'s concentration.
04

Effects of Removing H2S

Removing \( \mathrm{H}_{2} \mathrm{S} \) decreases the concentration of one of the products. Le Chatelier’s principle indicates that the equilibrium will shift toward the products, in this case releasing more \( \mathrm{NH}_{3} \) and \( \mathrm{H}_{2} \mathrm{S} \) to counterbalance the removal. Consequently, the pressure of \( \mathrm{NH}_{3} \) will increase as it shifts to produce more \( \mathrm{H}_{2} \mathrm{S} \).

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Key Concepts

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

Endothermic Reactions
Endothermic reactions are intriguing chemical processes that absorb heat from their surroundings. This is opposite to exothermic reactions, which release heat. In an endothermic reaction, like the decomposition of ammonium hydrogen sulfide (\(\mathrm{NH}_{4} \mathrm{HS}\)) into ammonia (\(\mathrm{NH}_{3}\)) and hydrogen sulfide (\(\mathrm{H}_{2} \mathrm{S}\)), energy is taken in as heat to break bonds in the reactant molecules.
Increasing the temperature in an endothermic reaction provides the additional heat required for the reaction to proceed further towards forming products. According to Le Chatelier's principle, this addition of heat will cause the equilibrium to shift towards the products side, favoring the formation of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\). This is because the system naturally wants to absorb the excess heat, balancing out the change imposed.
Chemical Equilibrium
Chemical equilibrium occurs when a reversible reaction achieves a state where the forward and reverse reactions happen at equal rates. At this point, the concentrations of reactants and products remain constant over time, though they are not necessarily equal.
For the decomposition reaction \(\mathrm{NH}_{4} \mathrm{HS} ightleftharpoons \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{S}\), equilibrium is an essential concept. Once equilibrium is reached, any disturbance like changes in concentration, temperature, or pressure will cause the system to adjust (or shift) to restore equilibrium, as predicted by Le Chatelier's principle.
  • Adding heat to an endothermic reaction shifts equilibrium towards products.
  • Adding reactants or removing products shifts the equilibrium to the right, favoring product formation.
  • Conversely, adding products or removing reactants shifts the equilibrium to the left, favoring reactant formation.
Decomposition Reactions
Decomposition reactions involve the breaking down of a compound into simpler substances. These reactions often require energy input, such as heat, light, or electricity, to proceed. In our case, the decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) results in the formation of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) gases.
Such reactions are crucial in various industrial and natural processes. The decomposition can be affected by the supply of energy and the physical conditions imposed on the system. In this case, increasing the temperature aids in overcoming the energy hurdle needed to break the bonds in \(\mathrm{NH}_{4} \mathrm{HS}\), hence pushing the reaction towards producing more \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\). Understanding the nature of decomposition reactions helps in translating theoretical knowledge into real-world applications.
Equilibrium Constant
The equilibrium constant (\(K_e\)) is a crucial aspect of understanding chemical equilibria. It quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible reaction at a given temperature.In the decomposition reaction \(\mathrm{NH}_{4} \mathrm{HS} \rightleftharpoons \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{S}\), the equilibrium constant can be expressed as:\[K_e = \frac{[\mathrm{NH}_{3}][\mathrm{H}_{2} \mathrm{S}]}{[\mathrm{NH}_{4} \mathrm{HS}]}\]It remains constant unless the temperature changes, which might increase the rate of product formation in an endothermic reaction. However, the equilibrium constant does not alter with changes in the concentration of solids or an inert gas's addition, as they do not appear in the equilibrium expression. Keeping the equilibrium constant in mind is vital for predicting which way a reaction will shift when a system is disturbed, as encapsulated in Le Chatelier's principle.

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Most popular questions from this chapter

A mixture of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) is placed in a reaction flask: \([\mathrm{CO}]=0.0102 \mathrm{mol} / \mathrm{L}\) and \(\left[\mathrm{Cl}_{2}\right]=0.00609 \mathrm{mol} / \mathrm{L}\) When the reaction $$ \mathbf{C O}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \operatorname{COCl}_{2}(\mathrm{g}) $$ has come to equilibrium at \(600 \mathrm{K},\left[\mathrm{Cl}_{2}\right]=\) \(0.00301 \mathrm{mol} / \mathrm{L}\) (a) Calculate the concentrations of \(\mathrm{CO}\) and \(\mathrm{COCl}_{2}\) at equilibrium. (b) Calculate \(K_{c}\).

The equilibrium constant for the reaction $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ is \(1.7 \times 10^{-3}\) at \(2300 K\) (a) What is \(K\) for the reaction when written as follows? $$ 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g}) $$ (b) What is \(K\) for the following reaction? $$ 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

Consider the following equilibrium: \(\operatorname{cosr}_{2}(\mathrm{g}) \rightleftarrows \mathrm{CO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.190\) at \(73^{\circ} \mathrm{C}\) (a) A 1.0 mol sample of \(\mathrm{COBr}_{2}\) is transferred to a \(2.0-\mathrm{L}\) flask and heated until equilibrium is attained. Calculate the equilibrium concentrations of each species. (b) The volume of the container is decreased to \(1.0 \mathrm{L}\) and the system allowed to return to equilibrium. Calculate the new equilibrium concentrations. (Hint. The calculation will be easier if you view this as a new problem with the new initial concentration of \(\left.\mathrm{COBr}_{2} \text { of } 1.0 \mathrm{mol} / \mathrm{L}_{.}\right)\)

Write equilibrium constant expressions for the following reactions. For gases, use either pressures orl concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{g})\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftarrows 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)

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

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