`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 calculating the percentage of α-(D) glucose at equilibrium when it converts to β-(D) glucose with an equilibrium constant (K) of 1.8, we can follow these steps:

### Step-by-Step Solution:

1. **Define Initial Concentrations**:
- Let the initial concentration of α-(D) glucose be 1 (1 mole).
- The initial concentration of β-(D) glucose is 0.

2. **Define Change at Equilibrium**:
- At equilibrium, let the concentration of α-(D) glucose be \( \alpha \).
- Therefore, the concentration of β-(D) glucose at equilibrium will be \( 1 - \alpha \).

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

4. **Solve for α**:
- Cross-multiply to solve for \( \alpha \):
\[ \alpha = 1.8(1 - \alpha) \]
\[ \alpha = 1.8 - 1.8\alpha \]
\[ \alpha + 1.8\alpha = 1.8 \]
\[ 2.8\alpha = 1.8 \]
\[ \alpha = \frac{1.8}{2.8} \approx 0.642 \]

5. **Calculate the Concentration of β-(D) Glucose**:
- The concentration of β-(D) glucose at equilibrium is:
\[ 1 - \alpha = 1 - 0.642 = 0.358 \]

6. **Calculate the Percentage of α-(D) Glucose**:
- The percentage of α-(D) glucose at equilibrium is given by:
\[ \text{Percentage of α-(D) glucose} = \alpha \times 100 = 0.642 \times 100 \approx 64.2\% \]

7. **Final Answer**:
- The percentage of α-(D) glucose at equilibrium is approximately **64.2%**.