Question

Question: The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

C₁₂ H₂₂ O₁₁(aq) + H₂O(l)⟶C₆ H₁₂ O₆ (aq) + C₆ H₁₂ O₆ (aq)

Rate = k[C₁₂H₂₂O₁₁]

In neutral solution, k = 2.1 × 10⁻¹¹/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C₆H₁₂O₆, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 10⁵ at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

Short answer

The Concentration of

Glucose = \(0.150{\rm{M}}\)

Fructose = \(0.150{\rm{M}}\)

Sucrose = \(1.65 \times {10^{ - 7}}{\rm{M}}\).

Step-by-step solution

Definition

For first-order reactions, the equationln[A] = -kt + ln[A]₀ is similar to that of a straight line (y = mx + c) with slope -k. This line can be graphically plotted as follows. Thus, the graph for ln[A] v/s t for a first-order reaction is a straight line with slope -k.

Hydrolysis of sugar to glucose and fructose

\({{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}({\rm{aq}}) + {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}({\rm{aq}})\)

The Equillibrium constant \({K_c} = 1.36 \times {10^5}\)

The Concentration of Sucrose is \(0.150{\rm{M}}\)

We have to find the concentrations of glucose, fructose and sucrose at equilibrium

First, we assume that all of the sucrose is decomposed into fructose and glucose

Therefore, the initial concentration of

Sucrose \(0{\rm{M}}\)

Glucose \(0.150{\rm{M}}\)

Fructose \(0.150{\rm{M}}\)

\({{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}}({\rm{aq}}) + {{\rm{H}}_2}{\rm{O}}({\rm{l}}) \to {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}({\rm{aq}}) + {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}({\rm{aq}})\)

Initial (M)	\(0\)		\(0.150\)	\(00.150\)
Change (M)	\( + {\rm{x}}\)		\( - {\rm{X}}\)	\( - {\rm{X}}\)
Equillibrium(M)	\({\rm{X}}\)		\(0.150 - x\)	\(0.150 - x\)

Find the value of X

\({K_c} = \frac{{({\rm{\;fructose\;}}) \times ({\rm{\;glucose\;}})}}{{({\rm{\;sucrose\;}})}}\),

\(1.36 \times {10^5} = \frac{{(0.150 - x) \times (0.150 - x)}}{x}\)

Since \({{\rm{K}}_ - }c\)is larger than \({10^4}\)

We will assume that \(0.150 - {\rm{x}} \approx 0.150\)

\(\begin{array}{*{20}{c}}{1.36 \times {{10}^5} = \frac{{(0.150) \times (0.150)}}{x}}\\{x = 1.65 \times {{10}^{ - 7}}{\rm{M}}}\end{array}\)

Therefore, the concentrations of glucose, fructose and sucrose at equilibrium is

Glucose = \(0.150{\rm{M}} - 1.65 \times {10^{ - 7}}{\rm{M}} = 0.150{\rm{M}}\)

Fructose = \(0.150{\rm{M}} - 1.65 \times {10^{ - 7}}{\rm{M}} = 0.150{\rm{M}}\)

Sucrose = \(x = 1.65 \times {10^{ - 7}}{\text{M}}\)