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 will calculate the change in entropy for the system, the surroundings, and the universe step by step.

### 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:

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

Where:
- ΔH_vap = Enthalpy of vaporization = 349 kJ/mol = 349,000 J/mol (conversion to Joules)
- T = Temperature at which the vaporization occurs = 3500 K

Now substituting the values:

\[ \Delta S_{\text{system}} = \frac{349,000 \, \text{J/mol}}{3500 \, \text{K}} = 99.71 \, \text{J/K/mol} \]

### Step 2: Calculate ΔS_surrounding

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

\[ \Delta S_{\text{surrounding}} = -\frac{\Delta H_{\text{vap}}}{T} \]

Where:
- ΔH_vap = 349,000 J/mol
- T = Temperature of the surroundings = 3500 K

Now substituting the values:

\[ \Delta S_{\text{surrounding}} = -\frac{349,000 \, \text{J/mol}}{3500 \, \text{K}} = -99.71 \, \text{J/K/mol} \]

### Step 3: Calculate ΔS_universe

The change in entropy for the universe (ΔS_universe) is the sum of the changes in entropy for the system and the surroundings:

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

Substituting the values we calculated:

\[ \Delta S_{\text{universe}} = 99.71 \, \text{J/K/mol} - 99.71 \, \text{J/K/mol} = 0 \, \text{J/K/mol} \]

### Summary of Results

- ΔS_system = 99.71 J/K/mol
- ΔS_surrounding = -99.71 J/K/mol
- ΔS_universe = 0 J/K/mol