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Chapter 9: Problem 20

Write the equilibrium expressions \(K\) for the following reactions. (a) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{NO}_{2}(\mathrm{~g})\) (b) \(\mathrm{SbCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{SbCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \Rightarrow \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{~g})\)

Short Answer

Expert verified
\( K_a = \frac{[NO_{2}]^2}{[NO]^2[O_{2}]} \), \( K_b = \frac{[SbCl_{3}][Cl_{2}]}{[SbCl_{5}]} \), \( K_c = \frac{[N_{2}H_{4}]}{[N_{2}][H_{2}]^2} \) (assuming reaction (c) reaches equilibrium).

Step by step solution

01

Identify Reactants and Products for Reaction (a)

For the first reaction, identify the reactants as 2 NO(g) and O2(g), and the product as 2 NO2(g).
02

Write the Equilibrium Expression for Reaction (a)

The equilibrium expression is written as the product of the concentrations of the products raised to the power of their coefficients, divided by the product of the concentrations of the reactants raised to the power of their coefficients. For reaction (a), the equilibrium constant expression is: \( K = \frac{[NO_{2}]^2}{[NO]^2[O_{2}]} \).
03

Identify Reactants and Products for Reaction (b)

For the second reaction, the reactants are SbCl5(g), and the products are SbCl3(g) and Cl2(g).
04

Write the Equilibrium Expression for Reaction (b)

Following the same rule, the equilibrium expression for reaction (b) is: \( K = \frac{[SbCl_{3}][Cl_{2}]}{[SbCl_{5}]} \).
05

Identify Reactants and Products for Reaction (c)

For the third reaction, the reactants are N2(g) and H2(g), and the product is N2H4(g).
06

Write the Equilibrium Expression for Reaction (c)

The equilibrium expression for reaction (c) is slightly different because the reaction is not at equilibrium; it is indicated by a one-way arrow. However, if it were at equilibrium, the expression would be: \( K = \frac{[N_{2}H_{4}]}{[N_{2}][H_{2}]^2} \).

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Key Concepts

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

Understanding Chemical Equilibrium
Chemical equilibrium is a foundational concept in chemistry that occurs when the rates of the forward and reverse reactions in a system are equal, leading to no net change in the concentration of reactants and products over time. Imagine a busy crossroad where an equal number of cars are entering and leaving the intersection; this is analogous to a chemical system at equilibrium.

  • At equilibrium, the system's properties like concentration become constant, though the reactions continue at a molecular level.
  • This state can be described by an equilibrium constant (\(K\)), which provides a way to quantify the balance between reactants and products.
  • The value of this constant is fixed for a given reaction at a specific temperature.
  • Changes in concentration, temperature, or pressure can shift the equilibrium position but do not change the value of the equilibrium constant unless the temperature changes.

By understanding equilibrium, students can predict how a reaction will behave under various conditions, which is crucial in fields such as pharmaceuticals, environmental science, and chemical engineering.
The Reaction Quotient
The reaction quotient (\(Q\)) is like a snapshot of a reaction that isn't at equilibrium. It's calculated in the same way as the equilibrium constant, using the same formula, but with the current concentrations of the reactants and products, regardless of whether the system is at equilibrium.

  • \(Q\) helps determine which direction a reaction will proceed to reach equilibrium: forward, reverse, or if it's already there.
  • If \(Q < K\), the reaction will proceed forward to produce more products.
  • If \(Q > K\), the reaction will go in reverse to produce more reactants.
  • When \(Q = K\) the system is at equilibrium.

By calculating \(Q\) and comparing it to \(K\), students gain insight into the current state of a reaction and what adjustments might be needed to achieve equilibrium.
Applying Le Chatelier's Principle
The famous Le Chatelier's principle is like the guidebook for navigating the shifts in equilibrium. It states that if an external stress is applied to a system at equilibrium, the system will adjust itself to minimize that stress.

  • Stresses include changes in concentration, pressure, and temperature.
  • Adding more reactant will push the equilibrium to produce more product, while removing a product will also lead to more product being formed.
  • Changing the pressure by changing the volume of a gas reaction will shift the equilibrium toward the side with a lower number of gas molecules.
  • Temperature changes can increase or decrease \(K\) depending on whether the reaction is endothermic or exothermic.

Le Chatelier's principle is central in industrial processes such as the Haber process for ammonia synthesis, where conditions are optimized to yield the most product. Understanding this principle allows students to predict the outcome of changing conditions on the position of equilibrium.

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Most popular questions from this chapter

Calculate the standard reaction frec cnergy for each of the following reactions: (a) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \neq 2 \mathrm{HI}(\mathrm{g}), K=160\) at \(500 \mathrm{~K}\) (b) \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}), K=47.9\) at \(400 \mathrm{~K}\)

Write the equilibrium expressions \(K_{c}\) for the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{COCl}(\mathrm{g})+\mathrm{Cl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Br}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HBr}(\mathrm{g})\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})=2 \mathrm{SO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

\( \mathrm{~A} 25.0 \mathrm{~g}\) sample of ammonium carbamate, \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)\), was placed in an evacuated \(0.250-\mathrm{L}\). flask and kept at \(25^{\circ} \mathrm{C}\). At equilibrium, the flask contained \(17.4 \mathrm{mg}\) of \(\mathrm{CO}_{2}\). What is the value of \(K_{c}\) for the decomposition of ammonium carbamate into ammonia and carbon dioxide? The reaction is \(\mathrm{NH}_{4}\left(\mathrm{NH}_{2} \mathrm{CO}_{2}\right)(\mathrm{s})=2 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) .\)

The temperature dependence of the equilibrium constant of the reaction \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})\), which makes an important contribution to atmospheric nitrogen oxides, can be expressed as \(\ln K=2.5-\) \((21700 \mathrm{~K}) \cdot T^{-1}\). What is the standard enthalpy of the forward reaction?

Predict whether each of the following equilibria will shift toward products or reactants with a temperature increase. (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g)=2 \mathrm{NO}_{2}(g), \Delta H^{\circ}=+57 \mathrm{~kJ}\) (b) \(\mathrm{X}_{2}(\mathrm{~g})=2 \mathrm{X}(\mathrm{g})\), where \(\mathrm{X}\) is a halogen (c) \(\mathrm{Ni}(\mathrm{s})+4 \mathrm{CO}(\mathrm{g})=\mathrm{Ni}(\mathrm{CO})_{4}(\mathrm{~g}), \Delta \mathrm{H}^{+}=-161 \mathrm{~kJ}\) (d) \(\mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{NH}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\), \(\Delta H^{\circ}=-90 \mathrm{~kJ}\)
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