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?

Short answer

The equilibrium constant for the reaction is 1.16 and at equilibrium, 53.7% of fructose is converted to glucose.

Step-by-step solution

Calculation of Change in Concentration

The initial concentration of fructose, \(C_{f-initial} = 0.244 M\) and the concentration at equilibrium, \(C_{f-equilibrium} = 0.113 M\). So, the change in concentration \(\Delta C_f = C_{f-initial} - C_{f-equilibrium} = 0.244 M - 0.113 M = 0.131 M\). Since fructose is converting to glucose, this is also the concentration of glucose at equilibrium.

Calculation of the Equilibrium Constant

The equilibrium constant \(K_c\) for the reaction is given by the ratio of the concentration of products \(glucose\) to reactants \(fructose\) at equilibrium. Here the stoichiometry is one-to-one, so we have \(K_c = [Glucose] / [Fructose]\) where the square brackets denote concentrations at equilibrium. Substituting the values, we get \(K_c = 0.131 / 0.113 = 1.16\).

Calculation of Percentage Conversion

The percentage conversion of fructose to glucose is the ratio of the change in fructose concentration to the initial concentration, multiplied by 100. So we get Percent Conversion = \(\Delta C_f / C_{f-initial} * 100 = 0.131 / 0.244 * 100 = 53.7%\). This means that 53.7% of the fructose has been converted into glucose at equilibrium.

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the concentrations of reactants and products remain constant over time. This occurs when the forward reaction, which converts reactants into products, happens at the same rate as the reverse reaction, converting the products back into reactants. At equilibrium, the reaction does not stop but continues to occur in both directions. This results in stable concentrations of the different chemical species involved. For the equilibrium involving fructose converting to glucose, as stated in the problem, we reach equilibrium when the concentration of fructose stops changing because the rate at which it turns into glucose is equal to the rate at which glucose turns back into fructose. It's like a seesaw perfectly balanced in the middle. The equilibrium constant, represented as \( K_c \), is a crucial concept here, as it gives us a quantitative measure of the position of equilibrium. A \( K_c \) greater than 1, as in the problem where \( K_c = 1.16 \), suggests that at equilibrium, the products (glucose) are favored over the reactants (fructose). Understanding this balance helps chemists predict the outcomes of chemical reactions in terms of amounts of products and reactants at equilibrium.

Exploring Reaction Kinetics

Reaction kinetics involves the study of the rates at which chemical processes occur. It helps us understand how various conditions affect the speed of a reaction. In the reaction between fructose and glucose, reaction kinetics would look at how quickly fructose is converted to glucose and vice versa. Several factors can influence these rates including temperature, concentration of reactants, and catalysts. Higher temperatures generally increase reaction rates as they provide reactant molecules with more energy, leading to more successful collisions. Reaction kinetics gives insight into the time it takes to reach equilibrium and helps in determining how the equilibrium constant links with reaction rates. For simple reactions where the forward and reverse processes are similar, the rate constants for forward and reverse reactions can be related to the equilibrium constant \( K_c \). This direct relationship indicates that the initial speed of the reaction contributes to the final balance of the reaction at equilibrium. Thus, understanding kinetics can help in manipulating reaction conditions to achieve desired levels of product or reactant more quickly.

Percentage Conversion and Its Implications

The concept of percentage conversion gives useful insight into how much of a reactant has been transformed into a product in a chemical reaction. It is represented as the percentage of the reactant's initial concentration that undergoes conversion to form products. In the context of the fructose-glucose equilibrium provided in the exercise, a percentage conversion of 53.7% means over half of the original fructose has turned into glucose by the time equilibrium is reached. To calculate the percentage conversion mathematically, you divide the change in concentration of the reactant by its initial concentration, then multiply by 100. This simple calculation provides a clear measure of the extent of a chemical reaction. It's a practical metric for understanding reaction efficiency in lab settings and industrial processes, allowing chemists to assess how effective a reaction is at converting reactants into desirable products. A high percentage conversion indicates an efficient reaction where a large portion of reactants is used up, while a low conversion might suggest the need for improved reaction conditions to yield more product.