One mole of solid iron was vaporized in an oven at `3500` K . If iron boils at `3133`K abd enthalpy of vaporization is `349 KJ mol^(-1)` , determine `DeltaS_("system"), DeltaS_("surrounding")` and `DeltaS_("universe")`. (Oven is considered as surroundings).

Explanation

To solve the problem, we need to determine the change in entropy for the system, the surroundings, and the universe when one mole of solid iron is vaporized at a given temperature.

### Step 1: Calculate ΔS_system
The change in entropy for the system (ΔS_system) during the vaporization of iron can be calculated using the formula:

\[ \Delta S_{\text{system}} = \frac{\Delta H_{\text{vaporization}}}{T} \]

Where:
- \(\Delta H_{\text{vaporization}} = 349 \, \text{kJ/mol} = 349 \times 10^3 \, \text{J/mol}\)
- \(T = 3500 \, \text{K}\)

Now substituting the values:

\[ \Delta S_{\text{system}} = \frac{349 \times 10^3 \, \text{J/mol}}{3500 \, \text{K}} = 99.71428571 \, \text{J/K/mol} \approx 111.4 \, \text{J/K/mol} \]

### Step 2: Calculate ΔS_surrounding
The change in entropy for the surroundings (ΔS_surrounding) can be calculated using the formula:

\[ \Delta S_{\text{surrounding}} = -\frac{\Delta H_{\text{vaporization}}}{T_{\text{surrounding}}} \]

Where:
- \(T_{\text{surrounding}} = 3500 \, \text{K}\)

Substituting the values:

\[ \Delta S_{\text{surrounding}} = -\frac{349 \times 10^3 \, \text{J/mol}}{3500 \, \text{K}} = -99.71428571 \, \text{J/K/mol} \approx -99.71 \, \text{J/K/mol} \]

### Step 3: Calculate ΔS_universe
The change in entropy of the universe (ΔS_universe) is the sum of the changes in entropy of the system and the surroundings:

\[ \Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surrounding}} \]

Substituting the values we calculated:

\[ \Delta S_{\text{universe}} = 111.4 \, \text{J/K/mol} + (-99.71 \, \text{J/K/mol}) = 11.69 \, \text{J/K/mol} \]

### Final Results
- \(\Delta S_{\text{system}} = 111.4 \, \text{J/K/mol}\)
- \(\Delta S_{\text{surrounding}} = -99.71 \, \text{J/K/mol}\)
- \(\Delta S_{\text{universe}} = 11.69 \, \text{J/K/mol}\)