Question

2.) when dissolved in water, glucose (corn sugar) and fructose (fruit sugar) exist in equilibrium as follows: fructose ⇌ glucose a chemist prepared a 0.244 m fructose solution at 25°c. at equilibrium, it was found that its concentration had decreased to 0.311 m. calculate the equilibrium constant for the reaction.

Explanation:

Step1: Calculate the change in fructose concentration

The initial concentration of fructose, $[Fructose]_{initial}=0.244\ M$. The equilibrium concentration of fructose, $[Fructose]_{eq}=0.111\ M$ (assuming there is a typo and it should be $0.111\ M$ instead of $0.311\ M$ as the concentration cannot increase if it is decreasing to reach equilibrium). The change in fructose concentration, $\Delta[Fructose]=[Fructose]_{initial}-[Fructose]_{eq}=0.244 - 0.111=0.133\ M$

Step2: Determine the equilibrium concentration of glucose

From the reaction $Fructose
ightleftharpoons Glucose$, the stoichiometric ratio is $1:1$. So, $[Glucose]_{eq}=\Delta[Fructose] = 0.133\ M$

Step3: Calculate the equilibrium constant

The equilibrium constant expression for the reaction $K=\frac{[Glucose]_{eq}}{[Fructose]_{eq}}$. Substitute the values: $K=\frac{0.133}{0.111}\approx1.20$

Answer:

The equilibrium constant for the reaction is approximately $1.20$