Question

Do the following reactions favor reactants or products at equilibrium? Give relative concentrations at equilibrium. (a) Sucrose \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows\) Glucose \((a q)+\) Fructose \((a q) \quad K=1.4 \times 10^{5}\) (b) \(\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftarrows \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q) \quad K=1.6 \times 10^{-5}\) (c) \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \rightleftarrows 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \quad K(\) at \(1000 \mathrm{~K})=24.2\)

Short answer

(a) Products favored; (b) Reactants favored; (c) Products favored.

Step-by-step solution

Understanding Equilibrium Constant

The equilibrium constant \( K \) gives us insight into the position of equilibrium. If \( K \) is much greater than 1, the reaction favors products at equilibrium. Conversely, if \( K \) is much less than 1, the reaction favors reactants.

Analyzing Reaction (a)

For the reaction \( \text{Sucrose (aq)} + \text{H}_2\text{O (l)} \rightleftharpoons \text{Glucose (aq)} + \text{Fructose (aq)} \), the equilibrium constant \( K = 1.4 \times 10^5 \). Since the value of \( K \) is much greater than 1, the reaction strongly favors the formation of glucose and fructose (products) at equilibrium.

Analyzing Reaction (b)

For the reaction \( \text{NH}_3 (aq) + \text{H}_2\text{O (l)} \rightleftharpoons \text{NH}_4^+ (aq) + \text{OH}^- (aq) \), the equilibrium constant \( K = 1.6 \times 10^{-5} \). This small value of \( K \) indicates that the reaction favors the reactants (NH3 and H2O) at equilibrium.

Analyzing Reaction (c)

For the reaction \( \text{Fe}_2\text{O}_3 (s) + 3 \text{CO (g)} \rightleftharpoons 2 \text{Fe (s)} + 3 \text{CO}_2 (g) \), the equilibrium constant \( K = 24.2 \) at 1000 K. Because \( K \) is greater than 1, the reaction favors the products (Fe and CO2) at equilibrium.

Conclusion and Relative Concentrations

For reactions (a) and (c), products are favored, so at equilibrium, the concentrations of products (glucose and fructose for reaction a, Fe and CO2 for reaction c) will be higher than reactants. For reaction (b), reactants are favored, so the concentrations of NH3 and H2O will be higher than the products NH4+ and OH-.

Key concepts

Equilibrium Constant

The equilibrium constant, symbolized as \( K \), is crucial in determining the position of equilibrium in a chemical reaction. Equilibrium represents the state in a reversible reaction where the forward and reverse reaction rates are equal. This doesn't mean the amounts of reactants and products are equal, rather, they reach a consistent ratio described by the equilibrium constant.
The expression for \( K \) arises from the concentrations of products over reactants when the system has reached equilibrium. For example, for a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is given by:\[K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}}\]Key points about \( K \):

• If \( K \gg 1 \), it indicates the products are favored at equilibrium.
• If \( K \ll 1 \), it shows the reactants are favored.
• The magnitude of \( K \) helps predict the reaction's favorability and possible shifts.

Reaction Favorability

Reaction favorability refers to whether a reaction will predominantly result in products or reactants at equilibrium. This can be inferred from the equilibrium constant value:

• A large \( K \) value (e.g., \( 1.4 \times 10^5 \) as in reaction (a)) suggests that almost all reactants are converted into products, signifying a product-favored reaction.
• A small \( K \) value (e.g., \( 1.6 \times 10^{-5} \) as in reaction (b)) indicates that the reactants remain largely unchanged, showing a reactant-favored process.
• Values of \( K \) near unity, while not seen in this exercise, suggest a balanced production of reactants and products at equilibrium.

Ultimately, understanding \( K \) offers a clear pathway to anticipate the direction and outcome of chemical reactions.

Concentration at Equilibrium

At equilibrium, the concentration of substances in a reaction reflects the extent to which reactants have been converted into products. This is inherently linked to the equilibrium constant \( K \), which offers an insight into the relative concentrations.
For reactions where \( K \) is significantly greater than 1, as in reaction (a), the equilibrium concentration of products (glucose and fructose) is much higher than that of the reactants (sucrose and water). Conversely, with a very small \( K \), as in reaction (b), the reactants (NH3 and H2O) have a significantly higher concentration than the products (NH4+ and OH-).
Therefore, knowing \( K \) helps in predicting:

• Which species will dominate the equilibrium mixture.
• How shifts in conditions (temperature, pressure) affect the concentrations of reactants and products.

This understanding is essential in fields like industrial chemistry, where reaction conditions are optimized to maximize yield and efficiency.