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

Question

The equilibrium constant for mutarotation α-D leftharpoons β-D Glucose is 1.8. What percentage of α form remains at equilibrium? (A) 35.7 (B) 64.3 (C) 55.6 (D) 44.4

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/9c4dc0ee8a47b7f21a38cf6c709421eb-.png" /> K=(α/1-α)=1.8 Solving α=0.642 1-α=0.358 Percent of α -D glucose remaining at equilibrium =35.8 %

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

35.7

64.3

55.6

44.4

$K = \frac{α}{1 - α} = 1.8$
Solving $α = 0.642$
$1 - α = 0.358$
Percent of $α - D$ glucose remaining at equilibrium
$= 35.8%$

Q. The equilibrium constant for mutarotation α−D⇌β−D Glucose is 1.8. What percentage of α form remains at equilibrium?