Question

Write equilibrium equations for the following reactions: (a) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g)\) (b) \(2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (c) \(2 \mathrm{BrF}_{5}(g) \rightleftarrows \mathrm{Br}_{2}(g)+5 \mathrm{~F}_{2}(g)\)

Short answer

(a) \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \); (b) \( K_c = \frac{[\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{H}_2\mathrm{S}]^2[\mathrm{O}_2]} \); (c) \( K_c = \frac{[\mathrm{Br}_2][\mathrm{F}_2]^5}{[\mathrm{BrF}_5]^2} \).

Step-by-step solution

Understand Equilibrium Expression

An equilibrium expression relates the concentrations of reactants and products at equilibrium. It is written as \( K_c = \frac{[products]^{coeff}}{[reactants]^{coeff}} \), where 'coeff' refers to the stoichiometric coefficients in the balanced equation.

Write Equation for Reaction (a)

For the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \), the equilibrium expression is \( K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \), using the stoichiometric coefficients as exponents.

Account for Pure Solids or Liquids in Reaction (b)

For \( 2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2\mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \), pure solids and pure liquids do not appear in the equilibrium expression. Therefore, \( K_c = \frac{[\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{H}_2\mathrm{S}]^2[\mathrm{O}_2]} \).

Write Equation for Reaction (c)

For the reaction \( 2 \mathrm{BrF}_{5}(g) \rightleftarrows \mathrm{Br}_{2}(g)+5 \mathrm{~F}_{2}(g) \), the equilibrium expression is \( K_c = \frac{[\mathrm{Br}_2][\mathrm{F}_2]^5}{[\mathrm{BrF}_5]^2} \), using the stoichiometric coefficients as exponents.

Key concepts

Equilibrium Expressions

Chemical equilibrium occurs when the forward and reverse reactions in a chemical process proceed at equal rates, meaning the concentrations of reactants and products remain constant over time. To represent this balanced state mathematically, we use equilibrium expressions. These expressions are crucial because they describe the ratio of the concentration of products to the concentration of reactants at equilibrium. The equilibrium constant, denoted as \( K_c \), is unique for each reaction at a specific temperature.An equilibrium expression is formulated from the chemical equation of a reaction. For example, in the reaction \( \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftarrows 2 \mathrm{NO}_{2}(g) \), the equilibrium expression is derived as:\[ K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]} \]Key points about equilibrium expressions:

• The concentrations of gases are expressed in terms of molarity, \([X]\).
• Coefficients in the balanced equation become the exponents in the equilibrium expression.
• Only the concentrations of gases and aqueous solutions are included; pure solids and liquids are omitted.

These expressions help predict the direction a reaction will shift under changing conditions, guiding chemists to manipulate reactions to obtain desired outcomes.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is derived from the balanced chemical equation, providing the proportions needed for different substances involved in the reaction. In equilibrium chemistry, stoichiometry defines the exponents in an equilibrium expression, as these are derived from the coefficients of the balanced equation.Using the reaction \( 2 \mathrm{BrF}_{5}(g) \rightleftarrows \mathrm{Br}_{2}(g)+5 \mathrm{F}_{2}(g) \), stoichiometry tells us that two moles of \( \mathrm{BrF}_5 \) decompose to form one mole of \( \mathrm{Br}_2 \) and five moles of \( \mathrm{F}_2 \). Therefore, the equilibrium expression is:\[ K_c = \frac{[\mathrm{Br}_2][\mathrm{F}_2]^5}{[\mathrm{BrF}_5]^2} \]Simplifying stoichiometry:

• Understand the mole ratio from the balanced equation.
• Apply these ratios to formulate precise equilibrium expressions.
• Use these relationships to calculate amounts of substances before, at, and after equilibrium is reached.

While stoichiometry seems daunting, it simply requires careful analysis of the balanced equation, guiding precise calculations in chemical processes.

Concentration of Reactants and Products

Concentration is a fundamental concept in chemical equilibrium, representing the amount of substance present in a given volume. Equilibrium expressions hinge on concentrations because they define how balanced a reaction is at any given time. The measures used are typically molarity (moles per liter).In any equilibrium analysis, determining the concentrations of all reactants and products is crucial. Take, for instance, the reaction \( 2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{O}_{2}(g) \rightleftarrows 2 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \). Despite having solid sulfur as a product, which is not included in the expression (as solids do not affect equilibrium), we account for all gases:\[ K_c = \frac{[\mathrm{H}_2\mathrm{O}]^2}{[\mathrm{H}_2\mathrm{S}]^2[\mathrm{O}_2]} \]Considerations for understanding concentration:

• Equilibrium concentrations are what appear in the expression, not initial concentrations.
• Concentrations change as reactions proceed towards equilibrium.
• Changes in external conditions, like pressure and temperature, can shift concentrations.

Understanding concentrations is vital for accurately predicting and controlling how chemical reactions behave at equilibrium.