The value of `DeltaH-DeltaE` for the following reaction at `27^(@)C` will be
`2NH_(3)(g)toN_(2)(g)+3H_(2)(g)`

Options

A. `8.314xx273xx(-2)`
B. `8.314xx300xx(-2)`
C. `8.314xx272xx(2)`
D. `8.314xx300xx(2)`

Answer

To find the value of \( \Delta H - \Delta E \) for the reaction \( 2NH_3(g) \rightarrow N_2(g) + 3H_2(g) \) at \( 27^\circ C \), we can use the relationship between enthalpy change (\( \Delta H \)) and internal energy change (\( \Delta E \)) given by the equation:

\[ \Delta H = \Delta E + \Delta N_g R T \]

Where:
- \( \Delta N_g \) = change in the number of moles of gas
- \( R \) = gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin

### Step 1: Calculate \( \Delta N_g \)

First, we need to determine \( \Delta N_g \), which is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.

- **Products**:
- \( N_2(g) \) = 1 mole
- \( H_2(g) \) = 3 moles
- **Total moles of products** = \( 1 + 3 = 4 \) moles

- **Reactants**:
- \( NH_3(g) \) = 2 moles
- **Total moles of reactants** = \( 2 \) moles

Now, we can calculate \( \Delta N_g \):

\[ \Delta N_g = \text{Total moles of products} - \text{Total moles of reactants} = 4 - 2 = 2 \]

### Step 2: Convert Temperature to Kelvin

Next, we convert the temperature from Celsius to Kelvin:

\[ T = 27^\circ C + 273.15 = 300.15 \approx 300 \text{ K} \]

### Step 3: Substitute Values into the Equation

Now we can substitute \( \Delta N_g \), \( R \), and \( T \) into the equation:

\[ \Delta H - \Delta E = \Delta N_g R T \]

Substituting the values:

\[ \Delta H - \Delta E = 2 \times 8.314 \, \text{J/(mol·K)} \times 300 \, \text{K} \]

### Step 4: Calculate the Result

Now, we perform the calculation:

\[ \Delta H - \Delta E = 2 \times 8.314 \times 300 = 4998.4 \, \text{J} \]

### Final Answer

Thus, the value of \( \Delta H - \Delta E \) for the reaction at \( 27^\circ C \) is approximately \( 5000 \, \text{J} \) or \( 5 \, \text{kJ} \).

---