Problem 37 - Chemistry

Question

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

Step-by-Step Solution

Answer

For the given reactions, the equilibrium constant expressions are: a) \(K = \frac{[\mathrm{H}_{2} \mathrm{O}]}{[\mathrm{NH}_{3}]^2 [\mathrm{CO}_{2}]}\) and \(K_{\mathrm{p}} = K\) b) No equilibrium, K and Kp expressions not applicable. c) No equilibrium, K and Kp expressions not applicable. d) \(K = \frac{[\mathrm{H}_{2} \mathrm{O}]}{[\mathrm{H}_{2}]}\) and \(K_{\mathrm{p}} = K\)

Step 1: a. Write expression for K and Kp for reaction a

For the reaction: \[2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\] We can find the equilibrium constant K as follows: \[K = \frac{[\mathrm{H}_{2} \mathrm{O}]}{[\mathrm{NH}_{3}]^2 [\mathrm{CO}_{2}]}\] Since reaction a involves only gas-phase species, we can write Kp expression as the same as K: \[K_{\mathrm{p}} = K\]

Step 2: b. Write expression for K and Kp for reaction b

For the reaction: \[2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\] There is no equilibrium reaction happening because the reaction goes to completion. The expression for K and Kp is not applicable.

Step 3: c. Write expression for K and Kp for reaction c

For the reaction: \[2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\] There is no equilibrium reaction happening because the reaction goes to completion. The expression for K and Kp is not applicable.

Step 4: d. Write expression for K and Kp for reaction d

For the reaction: \[\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\] We can find the equilibrium constant K as follows: \[K = \frac{[\mathrm{H}_{2} \mathrm{O}]}{[\mathrm{H}_{2}]}\] Since reaction d involves only gas-phase species, we can write Kp expression as the same as K: \[K_{\mathrm{p}} = K\]

Key Concepts

K and Kp expressions

Equilibrium constants, known as \( K \) and \( K_{\mathrm{p}} \), are essential for understanding chemical reactions at equilibrium. These constants help express the ratio of concentrations or partial pressures of products to reactants. For reactions involving gases, the equilibrium constant is often expressed as \( K_{\mathrm{p}} \), which uses partial pressures instead of concentrations. The basic idea is to see which side of the chemical equation is favored under equilibrium conditions.

The expression \( K = \frac{[\text{Products}]}{[\text{Reactants}]} \) uses the concentrations of species at equilibrium. For gas reactions, \( K_{\mathrm{p}} = \frac{(\text{Partial pressure of products})}{(\text{Partial pressure of reactants})} \) is employed.

• Products form the numerator.
• Reactants form the denominator.
• Solids and pure liquids are not included in these expressions.

It's important to balance the equation before forming \( K \) expressions as stoichiometry determines the exponents in the expression. Typically, use \( K \) when dealing with concentrations and \( K_{\mathrm{p}} \) for gaseous reactions in terms of partial pressures.

Chemical equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time. It's important to note that equilibrium doesn't mean reactants and products are in equal quantities; it simply means their rates of formation are equal, resulting in a dynamic balance.

There are several characteristics of chemical equilibrium:

• It is dynamic, meaning reactions continue to occur.
• The system is closed, with no net change in concentrations.
• The ratio of concentrations is constant, represented by the equilibrium constant \( K \).
• Temperature changes can shift the equilibrium position.

Understanding chemical equilibrium involves not just initial conditions, but how a system reaches a stable state where activities do not change further. Le Chatelier’s Principle is often applied to predict the effects of changing conditions on the equilibrium.

Phase states in reactions

Phase states in a chemical reaction play a vital role in determining how we calculate equilibrium expressions. Substances can exist in different phases, such as solids, liquids, and gases, all of which behave uniquely in reactions.

In equilibrium expressions, only species in the gas phase or in solution (aqueous) are typically included. This is because:

• The activities of pure solids and liquids are considered to be constant and thus omitted. This means they do not affect the equilibrium constant \( K \).
• In gas-phase reactions, the partial pressures of gaseous reactants and products are used in \( K_{\mathrm{p}} \).
• For reactions involving solids or liquids, the equilibrium is only influenced by aqueous or gaseous species.

Recognizing phase states can help simplify calculations and deepen understanding of the system balance. They indicate which species will significantly alter the equilibrium position when changes occur. By focusing on phases that impact \( K \) and \( K_{\mathrm{p}} \), one can streamline the process of analyzing and predicting chemical equilibria.