The heat of formation of `C_(2)H_(5)OH(l)` is `-66 kcal//"mole"`. The heat of combustion of `CH_(3)OCH_(3) (g)` is -348 kcal/mole. `DeltaH_(f)` for `H_(2)O` and `CO_(2)` are -68 kcal/mole respectively. Then, the `Delta H` for the isomerisation reaction `C_(2)H_(5)OH(l)rarrCH_(3)OCH_(3)(g)`,and `Delta E` for the same are at `T=25^(@)C`

Options

A. `Delta H=18` kcal/mole,`Delta E= 17.301`kcal/mole
B. `Delta H=22` kcal/mole,`Delta E=21.408` kcal/mole
C. `DeltaH=26` kcal/mole, `Delta E=25.709` kcal/mole
D. `Delta H=30` kcal/mole, `Delta E= 28.522` kcal/mole

Explanation

To solve the problem, we need to calculate the enthalpy change (ΔH) for the isomerization reaction of ethanol (C₂H₅OH) to dimethyl ether (CH₃OCH₃) and then calculate the change in internal energy (ΔE) for the same reaction.

### Step 1: Write the isomerization reaction
The isomerization reaction we are interested in is:
\[ \text{C}_2\text{H}_5\text{OH}(l) \rightarrow \text{CH}_3\text{OCH}_3(g) \]

### Step 2: Write the relevant thermochemical equations
1. Formation of ethanol:
\[ 3\text{C} + 3\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \quad \Delta H_f = -66 \text{ kcal/mol} \]

2. Combustion of dimethyl ether:
\[ \text{CH}_3\text{OCH}_3(g) + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \quad \Delta H_c = -348 \text{ kcal/mol} \]

3. Formation of water:
\[ \text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H_f = -68 \text{ kcal/mol} \]

4. Formation of carbon dioxide:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H_f = -94 \text{ kcal/mol} \]

### Step 3: Manipulate the equations to find ΔH for the isomerization
To find ΔH for the isomerization reaction, we can manipulate the above equations.

1. Multiply the formation of water equation (3) by 3:
\[ 3\text{H}_2 + \frac{3}{2}\text{O}_2 \rightarrow 3\text{H}_2\text{O} \quad \Delta H = 3 \times (-68) = -204 \text{ kcal} \]

2. Multiply the formation of CO₂ equation (4) by 2:
\[ 2\text{C} + 2\text{O}_2 \rightarrow 2\text{CO}_2 \quad \Delta H = 2 \times (-94) = -188 \text{ kcal} \]

3. Now, we add the modified equations (1) and (2) and subtract the combustion of ether (2) and the formation of ethanol (1):
\[ \Delta H = (-204) + (-188) - (-66) - (-348) \]
\[ = -204 - 188 + 66 + 348 \]
\[ = 22 \text{ kcal} \]

### Step 4: Calculate ΔE using the relation ΔH = ΔE + ΔN_gRT
Now we need to calculate ΔE. We use the relation:
\[ \Delta H = \Delta E + \Delta N_gRT \]

Where:
- ΔN_g = moles of gaseous products - moles of gaseous reactants
- In our reaction, we have 1 mole of gaseous product (CH₃OCH₃) and 0 moles of gaseous reactants (C₂H₅OH is a liquid), so:
\[ \Delta N_g = 1 - 0 = 1 \]

### Step 5: Substitute values to find ΔE
Using the universal gas constant R = 2 kcal/(mol·K) and T = 298 K:
\[ \Delta E = \Delta H - \Delta N_gRT \]
\[ = 22 \text{ kcal} - (1)(2 \text{ kcal/(mol·K)})(298 \text{ K})/1000 \]
\[ = 22 \text{ kcal} - 0.596 \text{ kcal} \]
\[ = 21.404 \text{ kcal} \approx 21.4 \text{ kcal} \]

### Final Results
- ΔH for the isomerization reaction is **22 kcal/mol**.
- ΔE for the isomerization reaction is **21.4 kcal/mol**.