Problem 86 - Chemistry

Question

When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 M\) fructose solution at \(25^{\circ} \mathrm{C}\). At equilibrium, it was found that its concentration had decreased to \(0.113 M .\) (a) Calculate the equilibrium constant for the reaction. (b) At equilibrium, what percentage of fructose was converted to glucose?

Step-by-Step Solution

Answer

So, (a) the equilibrium constant for the reaction is approximately 1.16, and (b) 53.7% of fructose was converted to glucose at equilibrium.

Step 1: Understanding the reaction and what happens at equilibrium

Fructose and glucose are in equilibrium in the solution. Therefore, the change in the concentration of fructose will be equal to the change in the concentration of glucose. When equilibrium is reached, the concentration of fructose has decreased from 0.244 M to 0.113 M, which mean the change in concentration is 0.244-0.113 = 0.131 M. We can conclude that 0.131 M of fructose has been converted into glucose.

Step 2: Calculating the equilibrium constant (K)

The equilibrium constant (K) is calculated by using the concentrations at equilibrium. It is the ratio of the concentration of the products to the concentration of the reactants. However, in this case, the reactant and the product is the same substance in different forms. The equilibrium constant (K) can be defined by the equilibrium concentrations as \[K = [glucose] / [fructose]\]. We can substitute the equilibrium concentrations into the equation to get \[K = 0.131 / 0.113\].

Step 3: Solve for the equilibrium constant (K)

K is calculated to be 1.16, approximately.

Step 4: Calculate the percentage of fructose transformed into glucose

The percentage of fructose converted to glucose can be worked out from the initial and final concentration. The change of concentration divided by the initial concentration will give us the fraction transformed. Multiplying it by 100 will give us the percentage. Therefore: \[Percentage \ transformation = (0.244 - 0.113) / 0.244 * 100\% \]

Step 5: Solve for the percentage of fructose transformed into glucose

Percentage transformation is calculated to be 53.7% approximately.

Key Concepts

Understanding Equilibrium

In chemistry, equilibrium refers to the state where the concentrations of reactants and products remain constant over time. This condition occurs when the rate of the forward reaction equals the rate of the reverse reaction. For our example, involving fructose and glucose, equilibrium is established when these two sugars exchange back and forth at equal rates in the solution.
When a reaction reaches equilibrium, no net change in concentration occurs. However, it is essential to understand that equilibrium doesn't mean the concentrations of the reactants and products are equal; instead, it means they remain steady. This balance is influenced by the specific conditions of the reaction, such as temperature and pressure, which can affect whether more fructose or glucose predominates in a mixture.
Equilibrium is a dynamic process, characterized by:

• No further change in concentration of reactants and products.
• Continual motion of molecules in both directions.
• Steady ratios defining the equilibrium constant, symbolized as \( K \).

Understanding this concept lays the foundation for calculating and analyzing the equilibrium constant.

Significance of Concentration

Concentration specifies the amount of a substance in a given volume. It is crucial because it determines how reactions occur at the molecular level. For our exercise, the changes in fructose and glucose concentrations help us understand the conversion process that occurs at equilibrium.
In the case of fructose and glucose, the initial concentration of fructose is known as 0.244 M, and at equilibrium, it decreased to 0.113 M. This decrease directly reflects the amount of fructose that has been converted into glucose, which allows for the calculation of the equilibrium constant.
The concept of concentration also aids in understanding the reaction extent and helps to:

• Determine the equilibrium position in relation to the reactants and products.
• Quantify the percentage conversion, giving insights into reaction efficiency.

Metrics of concentration are pivotal not only in reaching the calculations but also in experimental planning to achieve desired outcomes.

Role of Glucose in Equilibrium

Glucose plays a significant role in maintaining equilibrium due to its equilibrated presence with fructose in the solution. At equilibrium, the concentration of glucose on the product side arises from its conversion from fructose.
This transformation indicates that the two sugars are capable of interconverting naturally, and their concentrations at equilibrium directly affect the calculation of the equilibrium constant, \( K \), as shown by the ratio of glucose to fructose concentrations.
Understanding the behavior of glucose involves:

• Recognizing the chemical nature of glucose as an isomer of fructose.
• Observing its formation from fructose, thereby influencing the equilibrium state.
• Calculating the conversion percentage to determine how substantially fructose has converted into glucose, and vice versa, within the equilibrium framework.

These aspects illustrate glucose’s integral role in reaching and defining the equilibrium of this particular chemical reaction between sugars.