For the reaction `N_(2)+3H_(2)iff2NH_(3),DeltaH` is

Options

A. `DeltaE-2RT`
B. `DeltaE-RT`
C. `DeltaE+RT`
D. `DeltaE+2RT`

Answer

To find the change in enthalpy (ΔH) for the reaction \( N_2 + 3H_2 \iff 2NH_3 \), we can use the relation:

\[ \Delta H = \Delta E + \Delta N_G RT \]

where:
- \(\Delta E\) is the change in internal energy,
- \(\Delta N_G\) is the change in the number of moles of gas,
- \(R\) is the universal gas constant,
- \(T\) is the temperature in Kelvin.

### Step 1: Determine the number of gaseous moles of reactants and products.

For the given reaction:
- Reactants: \( N_2 + 3H_2 \) → Total moles of gas = \( 1 + 3 = 4 \) moles
- Products: \( 2NH_3 \) → Total moles of gas = \( 2 \) moles

### Step 2: Calculate the change in the number of moles of gas (\(\Delta N_G\)).

\[ \Delta N_G = \text{Moles of products} - \text{Moles of reactants} \]
\[ \Delta N_G = 2 - 4 = -2 \]

### Step 3: Substitute \(\Delta N_G\) into the equation for \(\Delta H\).

Now we can substitute \(\Delta N_G\) into the equation:

\[ \Delta H = \Delta E + (-2)RT \]
\[ \Delta H = \Delta E - 2RT \]

### Step 4: Conclusion.

The change in enthalpy (\(\Delta H\)) for the reaction can be expressed as:

\[ \Delta H = \Delta E - 2RT \]

This equation shows how the change in enthalpy is related to the change in internal energy and the change in the number of gaseous moles.