`alpha-(D)" glucose "hArr beta-(D)` glucose, equilibrium constant for this s 1.8. The percentage of `alpha-(D)` glucose at equilibrium is

Options

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

Answer

To solve the problem of finding the percentage of α-(D) glucose at equilibrium given the equilibrium constant (K) for the conversion of α-(D) glucose to β-(D) glucose is 1.8, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Initial Conditions:**
- Let the initial amount of α-(D) glucose be 1 mole.
- At the start, the amount of β-(D) glucose is 0 moles.

2. **Define the Change at Equilibrium:**
- Let α be the degree of dissociation of α-(D) glucose.
- At equilibrium, the amount of α-(D) glucose will be (1 - α) moles.
- The amount of β-(D) glucose formed will be α moles.

3. **Write the Expression for the Equilibrium Constant (K):**
- The equilibrium constant K for the reaction is given by:
\[ K = \frac{[\text{β-(D) glucose}]}{[\text{α-(D) glucose}]} \]
- At equilibrium, substituting the values:
\[ K = \frac{\alpha}{1 - \alpha} \]
- Given that K = 1.8, we can set up the equation:
\[ 1.8 = \frac{\alpha}{1 - \alpha} \]

4. **Solve for α:**
- Cross-multiplying gives:
\[ 1.8(1 - \alpha) = \alpha \]
- Expanding this:
\[ 1.8 - 1.8\alpha = \alpha \]
- Rearranging terms:
\[ 1.8 = \alpha + 1.8\alpha \]
\[ 1.8 = 2.8\alpha \]
- Solving for α:
\[ \alpha = \frac{1.8}{2.8} = 0.642857 \]

5. **Calculate the Percentage of α-(D) Glucose at Equilibrium:**
- The amount of α-(D) glucose at equilibrium is (1 - α):
\[ 1 - \alpha = 1 - 0.642857 = 0.357143 \]
- To find the percentage:
\[ \text{Percentage of α-(D) glucose} = (1 - \alpha) \times 100 = 0.357143 \times 100 = 35.7143\% \]

6. **Final Answer:**
- The percentage of α-(D) glucose at equilibrium is approximately **35.7%**.