When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be `(DeltaH_("vap")` (for water) `=40.8kJ"mol"^(-1)`)

Options

A. `4.08kJ, 10.93JK^(-1)`
B. `-4.08kJ, -40.8JK^(-1)`
C. `-4.04kJ,-10.93JK^(-1)`
D. `4.08kJ,40.8JK^(-1)`

Explanation

To solve the problem of calculating the enthalpy and entropy changes when 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, we can follow these steps:

### Step 1: Calculate the number of moles of steam
To find the number of moles, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
The molar mass of water (H₂O) is approximately 18 g/mol.

Given:
- Mass of steam = 1.8 g

Calculating the number of moles:
\[ \text{Number of moles} = \frac{1.8 \text{ g}}{18 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 2: Calculate the enthalpy change (ΔH)
The enthalpy change for the conversion of steam to water is the negative of the enthalpy of vaporization since we are condensing steam to water. The given enthalpy of vaporization (ΔH_vap) for water is 40.8 kJ/mol.

Thus, the enthalpy change for 0.1 moles is:
\[ \Delta H = -\Delta H_{\text{vap}} \times \text{number of moles} = -40.8 \text{ kJ/mol} \times 0.1 \text{ mol} = -4.08 \text{ kJ} \]

### Step 3: Calculate the entropy change (ΔS)
Using the relationship between ΔH and ΔS at equilibrium:
\[ \Delta G = \Delta H - T\Delta S \]
At equilibrium, ΔG = 0, therefore:
\[ 0 = \Delta H - T\Delta S \implies \Delta H = T\Delta S \]

Rearranging gives:
\[ \Delta S = \frac{\Delta H}{T} \]

Using the temperature at the normal boiling point of water, which is 100 °C or 373 K:
\[ \Delta S = \frac{-4.08 \text{ kJ}}{373 \text{ K}} = \frac{-4080 \text{ J}}{373 \text{ K}} \approx -10.94 \text{ J/K} \]

### Summary of Results
- Enthalpy change (ΔH) = -4.08 kJ
- Entropy change (ΔS) ≈ -10.94 J/K

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