Question

Write equilibrium constant expressions ( \(K\) ) for the following reactions: (a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons\) \(2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) (b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) (c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\)

Short answer

Question: Write the equilibrium constant expressions for the following reactions: a) \(2 \mathrm{NO}_{3}^{-}(a q)+8 \mathrm{H}^{+}(a q)+3 \mathrm{Cu}(s) \rightleftharpoons 2 \mathrm{NO}(g)+3 \mathrm{Cu}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)\) b) \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g)\rightleftharpoons2\mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) c) \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\) Answer: a) \(K = \frac{[\mathrm{NO}]^{2}[\mathrm{Cu}^{2+}]^{3}}{[\mathrm{NO}_{3}^{-}]^{2}[\mathrm{H}^{+}]^{8}}\) b) \(K = \frac{[\mathrm{SO}_{2}]^{2}}{[\mathrm{O}_{2}]^{3}}\) c) \(K = \frac{1}{[\mathrm{Ca}^{2+}][\mathrm{CO}_{3}{ }^{2-}]}\)

Step-by-step solution

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(2 \mathrm{NO}_{3}^{-}(a q)\), \(8 \mathrm{H}^{+}(a q)\), and \(3 \mathrm{Cu}(s)\). - Products: \(2 \mathrm{NO}(g)\), \(3 \mathrm{Cu}^{2+}(a q)\), and \(4 \mathrm{H}_{2} \mathrm{O}(l)\).

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{[\mathrm{NO}]^{2}[\mathrm{Cu}^{2+}]^{3}}{[\mathrm{NO}_{3}^{-}]^{2}[\mathrm{H}^{+}]^{8}}\) _b)# Equilibrium Constant Expression for Reaction (b): Here is the reaction: \(2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g)\rightleftharpoons2\mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g)\) Now, let's find the equilibrium constant expression by following the same steps:

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(2 \mathrm{PbS}(s)\) and \(3 \mathrm{O}_{2}(g)\). - Products: \(2\mathrm{PbO}(s)\) and \(2 \mathrm{SO}_{2}(g)\).

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{[\mathrm{SO}_{2}]^{2}}{[\mathrm{O}_{2}]^{3}}\) _c)# Equilibrium Constant Expression for Reaction (c): Here is the reaction: \(\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}{ }^{2-}(a q) \rightleftharpoons \mathrm{CaCO}_{3}(s)\) Now, let's find the equilibrium constant expression by following the same steps:

Identify reactants and products

From the balanced chemical equation, we have: - Reactants: \(\mathrm{Ca}^{2+}(a q)\) and \(\mathrm{CO}_{3}{ }^{2-}(a q)\). - Product: \(\mathrm{CaCO}_{3}(s)\).

Write the equilibrium expression

According to the definition of \(K\) explained earlier, the equilibrium expression for this reaction is: \(K = \frac{1}{[\mathrm{Ca}^{2+}][\mathrm{CO}_{3}{ }^{2-}]}\) Note that the solid product \(\mathrm{CaCO}_{3}(s)\) is not included in the equilibrium expression.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fascinating concept in chemistry where a chemical reaction and its reverse occur at the same rate, leading to a state where the concentrations of the reactants and products remain unchanged over time. This does not mean that the reaction has stopped; rather, the forward and reverse reactions are happening simultaneously at equal rates.
At equilibrium, even though reactants are still converted into products and vice versa, the overall composition remains constant. This balance can be disturbed by changing conditions such as temperature, pressure, or concentration, a principle known as Le Chatelier's Principle. Understanding equilibrium is crucial because it helps us predict how a system will react to different changes, which is vital in both industrial processes and natural systems.
In the context of equilibrium constant expressions, the stability of equilibrium is represented by the equilibrium constant ( K ), which quantifies the ratio of the concentration of products to reactants at equilibrium. A large value of K indicates that, at equilibrium, the system contains more products than reactants, while a small value suggests the opposite.

Reaction Stoichiometry

Stoichiometry is the field of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's vital for choosing the right proportions of reactants to use in a reaction and ensuring that the desired product is obtained efficiently.
When writing the equilibrium constant expressions, stoichiometry also plays a crucial role. For each balanced equation:

• The coefficients of reactants and products become powers in the expression of the equilibrium constant.
• We only include species in the gaseous or aqueous phase. Solids and liquids are excluded from the expression.

For example, in the reaction where lead sulfide (PbS) and oxygen (O_2) react to form lead oxide (PbO) and sulfur dioxide (SO_2) , the stoichiometry tells us that two moles of PbS react with three moles of O_2 . In the equilibrium expression, for every unit of SO_2 generated, three units of O_2 are consumed, emphasizing the balance maintained by the coefficients in the balanced equation.

Chemical Reactions

Chemical reactions are processes where substances, the reactants, are transformed into different substances, the products. These processes are essential for everything from industrial manufacturing to the metabolic pathways in our bodies.
In each chemical reaction, bonds between atoms in the reactants are broken, and new bonds are formed in the products. The type of reaction can vary widely:

• From simple combination reactions where two atoms form a molecule,
• To complex reactions like combustion or synthesis reactions involving numerous steps.

In the context of equilibrium reactions, chemical reactions are reversible, meaning they can proceed in both forward and backward directions. This reversibility is why we can establish an equilibrium state, as not all reactants are fully converted to products and vice versa. Understanding chemical reactions in terms of equilibrium helps chemists control reaction conditions to favor the production of desired products, whether they are components of medications, fuels, or everyday materials.