The enthalpy changes of some process are given below
`alpha-D "glucose"_((s)) + H_(2)O rarr alpha -D- "glucose"_((aq))` Heat of dissolution =10.84kJ
`beta- D "glucose"_((s)) + H_(2)O rarr beta- D- "glucose"_((aq))` Heat of dissolution = 4.68kJ
`alpha-D "glucose"_((aq)) rarr beta - D- "glucose"_((aq)` Heat of mutarotation `=-1.16kJ`
The `DeltaH^(0) "for " alpha - D "glucose"_((s)) rarr beta -D "glucose"_((s))` is

Answer

To find the enthalpy change (ΔH°) for the process where α-D-glucose (s) converts to β-D-glucose (s), we can use the given enthalpy changes of the dissolution and mutarotation processes.

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Enthalpy Changes:**
- Reaction 1:
\[ \alpha-D \text{ glucose}_{(s)} + H_2O \rightarrow \alpha-D \text{ glucose}_{(aq)} \quad \Delta H_1 = +10.84 \text{ kJ} \]
- Reaction 2:
\[ \beta-D \text{ glucose}_{(s)} + H_2O \rightarrow \beta-D \text{ glucose}_{(aq)} \quad \Delta H_2 = +4.68 \text{ kJ} \]
- Reaction 3:
\[ \alpha-D \text{ glucose}_{(aq)} \rightarrow \beta-D \text{ glucose}_{(aq)} \quad \Delta H_3 = -1.16 \text{ kJ} \]

2. **Set Up the Overall Reaction:**
We want to find ΔH° for the following reaction:
\[ \alpha-D \text{ glucose}_{(s)} \rightarrow \beta-D \text{ glucose}_{(s)} \]

3. **Combine the Reactions:**
To find the ΔH° for the overall reaction, we can manipulate the three reactions:
- Start with Reaction 1 (dissolving α-D glucose):
\[ \alpha-D \text{ glucose}_{(s)} + H_2O \rightarrow \alpha-D \text{ glucose}_{(aq)} \quad (+10.84 \text{ kJ}) \]
- Subtract Reaction 2 (dissolving β-D glucose):
\[ \beta-D \text{ glucose}_{(aq)} + H_2O \rightarrow \beta-D \text{ glucose}_{(s)} \quad (-4.68 \text{ kJ}) \]
- Add Reaction 3 (mutarotation):
\[ \alpha-D \text{ glucose}_{(aq)} \rightarrow \beta-D \text{ glucose}_{(aq)} \quad (-1.16 \text{ kJ}) \]

4. **Calculate the Total Enthalpy Change:**
The overall enthalpy change can be calculated as follows:
\[ \Delta H° = \Delta H_1 - \Delta H_2 + \Delta H_3 \]
Substituting the known values:
\[ \Delta H° = 10.84 \text{ kJ} - 4.68 \text{ kJ} - 1.16 \text{ kJ} \]
\[ \Delta H° = 10.84 - 4.68 - 1.16 = 5.00 \text{ kJ} \]

5. **Final Answer:**
The enthalpy change (ΔH°) for the conversion of α-D-glucose (s) to β-D-glucose (s) is:
\[ \Delta H° = 5.00 \text{ kJ} \]