Chapter 7, Problem 7.64P

7-64 As we shall see in Chapter 20, there are two forms of glucose, designated alpha and betawhich are in equilibrium in aqueous solution. The equilibrium constant for the reaction is 1.5 at 30°C.

(a) If you begin with a fresh 1.0 M solution of D-glucose in water, what will be its concentration when equilibrium is reached?

(b) Calculate the percentage of glucose and of glucose present at equilibrium in aqueous solution at 30°C.

Interpretation Introduction

(a)

Interpretation:

The equilibrium concentrations of $\alpha$ -D-glucose have to be determined by given its initial concentration and the value of equilibrium constant.

Concept Introduction:

$\alpha$ -D-glucose is present in equilibrium with $\alpha$ -D-glucose in solutions. The equilibrium constant for this conversion is 1.5 at 30⁰ C. The equilibrium concentration of $\alpha$ -D-glucose can be calculated by figuring out the change in concentration of $\alpha$ -D-glucose using the K value.

Answer to Problem 7.64P

${[\alpha \text{-D-Glucose}]}_{\text{equilibrium}}=0.40M$

Explanation of Solution

The given equilibrium reaction is,

$\alpha \text{-D-Glucose(}\text{a}q\text{)}\rightleftharpoons \beta \text{-D-Glucose(}\text{a}q\text{)}\text{K}\text{=}\text{1}\text{.5}$

The initial concentration of $\alpha$ -D-glucose = 1.0 M.

Setting up the equilibrium changes to calculate the equilibrium concentrations:

$\begin{array}{l}\alpha \text{-D-Glucose(}\text{a}q\text{)}\rightleftharpoons \beta \text{-D-Glucose(}\text{a}q\text{)}\text{K}\text{=}\text{1}\text{.5}\\ \text{Initial}\text{(M)}\text{1}\text{.0}\text{0}\\ \text{Change(M)}-\text{X}+\text{X}\\ \text{Equilibrium(M)}\text{(1}\text{.0}-\text{X})\text{X}\end{array}$

$K=\frac{[\beta \text{-D-Glucose}]}{[\alpha \text{-D-Glucose}]}$

$1.5=\frac{\text{X}}{(1.0-\text{X})}$

$\begin{array}{l}1.5(1.0-\text{X})=\text{X}\\ \text{X}=0.60M\end{array}$

Therefore, ${[\alpha \text{-D-Glucose}]}_{\text{equilibrium}}=1.0-0.60=0.40M$

Interpretation Introduction

(b)

Interpretation:

The percentages of $\alpha$ -D-glucose and $\alpha$ -D-glucose in the solution have to be calculated.

Concept Introduction:

$\alpha$ -D-glucose is present in equilibrium with $\alpha$ -D-glucose in solutions. The equilibrium constant for this conversion is 1.5 at 30⁰ C. The equilibrium concentration of $\alpha$ -D-glucose can be calculated by figuring out the change in concentration of $\alpha$ -D-glucose using the K value.

Answer to Problem 7.64P

Percentage of $\alpha$ -D-glucose at equilibrium in the solution = 40 %.

Percentage of $\alpha$ -D-glucose at equilibrium in the solution = 60 %.

Explanation of Solution

The given equilibrium reaction is,

$\alpha \text{-D-Glucose(}\text{a}q\text{)}\rightleftharpoons \beta \text{-D-Glucose(}\text{a}q\text{)}\text{K}\text{=}\text{1}\text{.5}$

The initial concentration of $\alpha$ -D-glucose = 1.0 M.

From the above calculations, concentration of $\alpha$ -D-glucose at equilibrium= 0.40 M.

Percentage of $\alpha$ -D-glucose at equilibrium in the solution = $\frac{0.40M}{1.0M}\times 100=40\%$

Percentage of $\alpha$ -D-glucose at equilibrium in the solution = $\frac{0.60M}{1.0M}\times 100=60\%$