Chemical EquilibriumMedium
Question
The equilibrium constant for the mutarotation, α-D–glucose$\rightleftharpoons$β–D–glucose is 1.8. What percent of the α-form remains under equilibrium?
Options
A.35.7
B.64.3
C.55.6
D.44.4
Solution
α – D – Glucose $\rightleftharpoons \beta$– D – Glucose
a 0
Equilibrium a – x x
$K = \frac{x}{a - x} = 1.8 \Rightarrow x = \frac{9}{14}a$
∴% of α – D – Glucose remained = $\frac{a - x}{a} \times 100 = 35.7\%$
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