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When dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose \(\rightleftharpoons\) glucose A chemist prepared a \(0.244 \mathrm{M}\) 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

Expert verified
(a) The equilibrium constant is approximately 1.159. (b) About 53.69% of fructose is converted to glucose at equilibrium.

Step by step solution

01

Determine Initial and Change in Concentration

Initially, the concentration of fructose was given as \(0.244 \text{ M}\). At equilibrium, the concentration of fructose is \(0.113 \text{ M}\). Therefore, the change in concentration of fructose is \(0.244 - 0.113 = 0.131 \text{ M}\).
02

Calculate Glucose Concentration

Since the change in concentration of fructose is \(0.131 \text{ M}\) and fructose converts to glucose in a 1:1 stoichiometry, the concentration of glucose at equilibrium will also be \(0.131 \text{ M}\).
03

Write the Equilibrium Expression

The equilibrium expression for the reaction \( ext{fructose} \rightleftharpoons ext{glucose} \) is given by:\[K_c = \frac{[\text{glucose}]}{[\text{fructose}]}\]
04

Substitute Equilibrium Concentrations into Expression

Substitute the equilibrium concentrations into the equilibrium expression:\[K_c = \frac{0.131}{0.113}\]
05

Calculate the Equilibrium Constant

Perform the calculation to find the equilibrium constant:\[K_c = \frac{0.131}{0.113} \approx 1.159\]
06

Calculate Percentage Converted to Glucose

Calculate the percentage of fructose converted to glucose using the initial concentration:\[\text{Percentage converted} = \left(\frac{0.131}{0.244}\right) \times 100 \approx 53.69\%\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
To comprehend chemical equilibrium, understanding the equilibrium constant is essential. When two reactants, such as fructose and glucose, are at equilibrium, the ratio of their concentrations becomes constant. This ratio is known as the equilibrium constant, denoted as \(K_c\). It represents how far a chemical reaction proceeds. In our example of fructose converting to glucose, we calculate \(K_c\) using \\[K_c = \frac{[\text{glucose}]}{[\text{fructose}]}\]. \At equilibrium, fructose concentration is \(0.113 \, \text{M}\) and glucose is \(0.131 \, \text{M}\). Substituting these into the formula, \(K_c\) becomes \(1.159\). This value helps chemists understand the balance of products to reactants, showing that more glucose forms from fructose. This constant provides insights into the direction and extent of the reaction under given conditions.
Concentration Changes
The concept of concentration change is crucial to quantify how much a reactant turns into a product. Initially, the solution has \(0.244 \, \text{M}\) fructose. Upon reaching equilibrium, only \(0.113 \, \text{M}\) fructose remains. The decrease, \(0.131 \, \text{M}\), reflects the amount converted to glucose. Understanding these changes clarifies a reaction’s progress:
  • Initial concentration: measures the starting substance.
  • Equilibrium concentration: reflects the substance amount when forward and reverse reactions balance.
  • Change in concentration: indicates how much reactant transformed to product.
These values provide a clear picture of how chemical equilibrium shifts, enabling precise calculation of equilibrium constants.
Stoichiometry
Stoichiometry is a core aspect of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In our exercise, fructose converts to glucose in a 1:1 ratio. This means for every mole of fructose that reacts, one mole of glucose is formed. This simple stoichiometric relationship allows us to directly equate the change in fructose concentration to the formed glucose concentration. It enables us to accurately determine the concentrations of other substances involved in the reaction by understanding their stoichiometric coefficients. In essence, stoichiometry aids in predicting the amounts of products generated from a given amount of reactants.
Percentage Conversion
Percentage conversion gives a numeric representation of how much reactant has been converted to product. For fructose turning into glucose, we calculate it to understand the extent of the conversion:The formula is \[\text{Percentage converted} = \left(\frac{\text{Change in concentration}}{\text{Initial concentration}}\right) \times 100\].Using the values from the exercise, we calculate \(\text{Percentage converted} = \left(\frac{0.131}{0.244}\right) \times 100 \approx 53.69\%\).This means over half of the fructose present initially turned into glucose. Knowing the percentage conversion helps assess the efficiency and completeness of a reaction. It is a handy metric for chemists to evaluate reaction conditions and optimize processes.

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Most popular questions from this chapter

The equilibrium constant \(K_{P}\) for the reaction $$2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)$$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

Define equilibrium. Give two examples of a dynamic equilibrium.

Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Consider the reaction $$\begin{aligned}2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons & 2 \mathrm{SO}_{3}(g) \\\\\Delta H^{\circ} &=-198.2 \mathrm{~kJ} / \mathrm{mol} \end{aligned}$$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2}\), and \(\mathrm{SO}_{3}\) at equilibrium if we were to (a) increase the temperature, (b) increase the pressure, (c) increase \(\mathrm{SO}_{2},\) (d) add a catalyst, (e) add helium at constant volume.

A sealed glass bulb contains a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases. When the bulb is heated from \(20^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C},\) what happens to these properties of the gases: (a) color, (b) pressure, (c) average molar mass, (d) degree of dissociation (from \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) ), (e) density? Assume that volume remains constant.

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