The standard entropy change for the reaction
`4Fe(s)+3O_(2)(g)to 2Fe_(2)O_(3)(s) " is" - 550 J K^(-1) ` at 298 K . [Given The standard enthalpy change for the reaction is -165 ` " kJ mol"^(-1)`] .the temperature in K at which the reaction attains equilibrium is _______ .

Explanation

To find the temperature at which the reaction attains equilibrium, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

At equilibrium, \(\Delta G = 0\). Therefore, we can set up the equation:

\[ 0 = \Delta H - T \Delta S \]

Rearranging this gives us:

\[ T = \frac{\Delta H}{\Delta S} \]

### Step 1: Convert \(\Delta H\) to the same units as \(\Delta S\)

Given:
- \(\Delta H = -165 \, \text{kJ/mol}\)
- \(\Delta S = -550 \, \text{J/K}\)

Since \(\Delta H\) is in kilojoules, we need to convert it to joules:

\[ \Delta H = -165 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -165000 \, \text{J/mol} \]

### Step 2: Substitute the values into the equation

Now, we can substitute \(\Delta H\) and \(\Delta S\) into the equation for \(T\):

\[ T = \frac{-165000 \, \text{J/mol}}{-550 \, \text{J/K}} \]

### Step 3: Calculate \(T\)

Now, perform the calculation:

\[ T = \frac{165000}{550} = 300 \, \text{K} \]

### Final Answer

The temperature at which the reaction attains equilibrium is **300 K**.

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