The equilibrium constant for mutarotation α -D Glucose leftharpoons β -D Glucose is 1.8 . What percentage of α form remains at equilibrium?

Correct answer: A. 35.7 α -D glucose leftharpoons β -D glucose array lcc Initial & 1 & 0 At equilibrium & 1-α & α array K= (α)/(1-α)= 1.8 Solving α=0.642 1-α=0.358 Percent of α -D glucose remaining at equilibrium =35.8 \% .

AIIMS 2009

Exam: AIIMS
Year: 2009
Subject: Chemistry
Chapter: Chemical Equilibrium

AIIMS 2009 · Chemistry · Chemical Equilibrium

The equilibrium constant for mutarotation α -D Glucose leftharpoons...

The equilibrium constant for mutarotation

\alpha

-D Glucose

\rightleftharpoons \beta

-D Glucose is 1.8 . What percentage of

\alpha

form remains at equilibrium?

Options

A $35.7$
B $64.3$
C $55.6$
D $44.4$

Correct answer

A. $35.7$

Step-by-step solution

\alpha

-D glucose

\rightleftharpoons \beta

-D glucose

$\begin{array}{lcc}\text{ Initial } & 1 & 0 \\ \text{ At equilibrium } & 1-\alpha & \alpha\end{array}

\frac{\alpha}{1-\alpha}=

1.8

Solving

\alpha=0.642 1-\alpha=0.358

Percent of

\alpha

-D glucose remaining at equilibrium

=35.8 \%

.

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