The standerd enthalpy of formation of FeO and `Fe_(2)O_(3)` is `-65` kcal `"mole"^(-1)` and `-197 " kcal" " mole"^(-1)` respectively . A mixture of two oxides containes FeO and `Fe_(2)O_(3)` in the mole ratio `2 : 1` . If by oxidation, it is changed into a `1 : 2` mole ratio mixture , How much of thermal energy will released per mole of the internal mixture ?

Options

A. `13.4 kcal//"mole"`
B. `14.6 kcal//"mole"`
C. `15.7 kcal//"mole"`
D. `16.8 kcal//"mole"`

Answer

To solve the problem, we need to determine how much thermal energy is released when the mixture of FeO and Fe2O3 changes from a 2:1 mole ratio to a 1:2 mole ratio. We will follow these steps:

### Step 1: Understand the initial and final states of the mixture
Initially, we have a mixture of FeO and Fe2O3 in a 2:1 mole ratio. Let's assume we have 2a moles of FeO and a moles of Fe2O3.

- Initial moles of FeO = 2a
- Initial moles of Fe2O3 = a

After oxidation, the mixture changes to a 1:2 mole ratio. We need to find the new amounts of FeO and Fe2O3.

### Step 2: Set up the equations for the final state
Let x be the number of moles of FeO that are oxidized to form Fe2O3. After oxidation, the moles of FeO and Fe2O3 will be:

- Final moles of FeO = 2a - x
- Final moles of Fe2O3 = a + x

According to the problem, the final ratio is 1:2, which gives us the equation:

\[ \frac{2a + x}{2a - x} = 1/2 \]

### Step 3: Solve for x
Cross-multiplying gives us:

\[ 2(2a + x) = 1(2a - x) \]

Expanding and simplifying:

\[ 4a + 2x = 2a - x \]

Rearranging gives:

\[ 3x = -2a \]

Thus,

\[ x = \frac{6a}{5} \]

### Step 4: Calculate the change in thermal energy
The change in thermal energy (ΔH) is given by the difference in the enthalpy of formation of the products and reactants. The standard enthalpy of formation values are:

- ΔHf (FeO) = -65 kcal/mol
- ΔHf (Fe2O3) = -197 kcal/mol

The reaction can be represented as:

\[ 2 \text{FeO} + \text{O}_2 \rightarrow 2 \text{Fe2O3} \]

The change in enthalpy for the reaction is:

\[ \Delta H = [\Delta Hf (\text{Fe2O3})] - [\Delta Hf (\text{FeO})] \]

Substituting the values:

\[ \Delta H = [-197 \, \text{kcal/mol}] - [2 \times (-65 \, \text{kcal/mol})] \]

Calculating this gives:

\[ \Delta H = -197 + 130 = -67 \, \text{kcal/mol} \]

### Step 5: Calculate the total energy released
Since x moles of FeO are oxidized, we need to calculate the energy released per mole of the internal mixture. The total energy released can be calculated as:

\[ \text{Energy released} = \Delta H \times \frac{x}{3} \]

Substituting the values:

\[ \text{Energy released} = -67 \times \frac{6a/5}{3} = -13.4 \, \text{kcal/mol} \]

### Final Answer
The thermal energy released per mole of the internal mixture is **-13.4 kcal/mol**.

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