Question

Write equilibrium constant \((K)\) expressions for the following reactions: (a) \(\mathrm{I}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{IF}_{5}(g)\) (b) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(l)\) (c) \(2 \mathrm{H}_{2} \mathrm{~S}+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{SO}_{2}(g)\) (d) \(\mathrm{SnO}_{2}(s)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Sn}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

Question: Write the equilibrium constant expressions for the following chemical reactions: a) I2(g) + 5 F2(g) ⇌ 2 IF5(g) b) CO(g) + 2 H2(g) ⇌ CH3OH(l) c) 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(l) + 2 SO2(g) d) SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(l) Answer: a) \(K = \frac{[\mathrm{IF}_{5}]^2}{[\mathrm{I}_{2}][\mathrm{F}_{2}]^5}\) b) \(K = \frac{[\mathrm{CH}_{3}\mathrm{OH(l)}]}{[\mathrm{CO(g)}][\mathrm{H}_{2}(g)]^2}\) c) \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2[\mathrm{SO}_{2}(g)]^2}{[\mathrm{H}_{2}\mathrm{S(g)}]^2[\mathrm{O}_{2}(g)]^3}\) d) \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2}{[\mathrm{H}_{2}(g)]^2}\)

Step-by-step solution

Identify the reactants and the products

The reactants in this reaction are I2(g) and F2(g), while the product is IF5(g).

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{IF}_{5}]^2}{[\mathrm{I}_{2}][\mathrm{F}_{2}]^5}\). #b) CO(g) + 2 H2(g) ⇌ CH3OH(l)#

Identify the reactants and the products

The reactants in this reaction are CO(g) and H2(g), while the product is CH3OH(l).

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{CH}_{3}\mathrm{OH(l)}]}{[\mathrm{CO(g)}][\mathrm{H}_{2}(g)]^2}\). Note that the concentration of the liquid CH3OH is considered as 1 in the reaction, because liquids and solids are considered to be in a separate phase that doesn't affect the equilibrium constant. #c) 2 H2S(g) + 3 O2(g) ⇌ 2 H2O(l) + 2 SO2(g)#

Identify the reactants and the products

The reactants in this reaction are H2S(g) and O2(g), while the products are H2O(l) and SO2(g).

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2[\mathrm{SO}_{2}(g)]^2}{[\mathrm{H}_{2}\mathrm{S(g)}]^2[\mathrm{O}_{2}(g)]^3}\). Again, note that the concentration of the liquid H2O is considered as 1 in the reaction since liquids do not change the equilibrium constant. #d) SnO2(s) + 2 H2(g) ⇌ Sn(s) + 2 H2O(l)#

Identify the reactants and the products

The reactants in this reaction are SnO2(s) and H2(g), while the products are Sn(s) and H2O(l).

Write the equilibrium constant expression

According to the reaction stoichiometry, the equilibrium constant expression for this reaction can be written as: \(K = \frac{[\mathrm{H}_{2}\mathrm{O(l)}]^2}{[\mathrm{H}_{2}(g)]^2}\). As before, the concentrations of the solid SnO2 and Sn, as well as the liquid H2O, are considered as 1 in the reaction.

Key concepts

Reaction Stoichiometry

In chemical reactions, reaction stoichiometry is crucial for determining the proportions in which reactants are consumed and products are formed. Stoichiometry is the part of chemistry that uses a balanced chemical equation to calculate the relationships between the quantities of reactants and products. In the equation

• \( \mathrm{I}_{2}(g)+5 \mathrm{~F}_{2}(g) \rightleftharpoons 2 \mathrm{IF}_{5}(g) \), the stoichiometric coefficients indicate that one molecule of \( \mathrm{I}_{2} \) reacts with five molecules of \( \mathrm{F}_{2} \) to form two molecules of \( \mathrm{IF}_{5} \).
• Similarly, in the expression \( \mathrm{CO} + 2 \mathrm{H}_{2} \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH} \), one molecule of carbon monoxide combines with two molecules of hydrogen gas to produce one molecule of liquid methanol.

The stoichiometric coefficients are essential because they allow chemists to understand the exact proportions required for reactions to proceed without excess reactants.

Chemical Reactions

Chemical reactions involve the conversion of reactants into products. They are represented by chemical equations that include the symbols and formulas of the involved substances. These equations must be balanced to reflect the law of conservation of mass, meaning that atoms are neither created nor destroyed, only rearranged.Understanding chemical reactions also leads to exploring equilibrium states. For instance, in the reaction \( 2 \mathrm{H}_{2} \mathrm{~S} + 3 \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O} + 2 \mathrm{SO}_{2} \), gaseous hydrogen sulfide and oxygen yield water and sulfur dioxide. The equilibrium constant \( K \) helps predict the extent of the reaction and how far the reaction will proceed before reaching equilibrium.

• At equilibrium, the rate at which reactants are converted into products equals the rate at which products revert to reactants.
• This state allows for the calculation of \( K \), where its magnitude indicates whether products or reactants are favored at equilibrium.

The ability to write and analyze equilibrium expressions using reactions' stoichiometry enables chemists to predict and interpret the behavior of chemical systems.

Phase Equilibrium

Phase equilibrium plays an essential role in determining how substances in different phases, such as solids, liquids, and gases, interact in a system. When considering equilibrium constant expressions, it's important to recognize that pure solids and liquids are omitted from \( K \) calculations. This is because their concentrations are constant and do not change the state of equilibrium. For instance, in the reaction \( \mathrm{SnO}_{2}(s) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Sn}(s) + 2 \mathrm{H}_{2}\mathrm{O}(l) \), both the solid \( \mathrm{SnO}_{2} \) and the liquid \( \mathrm{H}_{2}\mathrm{O} \) are not included in the equilibrium constant expression. Here, only the gaseous \( \mathrm{H}_{2} \) is considered, resulting in the \( K \) being equal to \[\frac{[\mathrm{H}_{2}\mathrm{O}]^2}{[\mathrm{H}_{2}]^2}\] without including the solids and liquids.Understanding phase equilibrium in reactions allows one to identify which substances' concentrations affect the system at equilibrium, thus enabling precise control over the reaction conditions for desired outcomes.