For the reaction
`N_(2) + 3 H_(2) = 2 NH_(3)`,

Options

A. `Delta U + 2 RT`
B. `Delta U - 2RT`
C. `Delta U + RT`
D. `Delta U - RT`

Explanation

To solve the problem regarding the reaction \( N_2 + 3 H_2 \rightarrow 2 NH_3 \) and to express \( \Delta H \) in terms of \( \Delta U \) and \( RT \), we can follow these steps:

### Step 1: Understand the relationship between \( \Delta H \) and \( \Delta U \)

The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta U \)) is given by the equation:

\[ \Delta H = \Delta U + \Delta n_g RT \]

where \( \Delta n_g \) is the change in the number of moles of gas during the reaction, and \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

### Step 2: Calculate \( \Delta n_g \)

To find \( \Delta n_g \), we need to calculate the total number of moles of gaseous products and subtract the total number of moles of gaseous reactants.

- On the **reactant side**:
- \( N_2 \) contributes 1 mole.
- \( 3 H_2 \) contributes 3 moles.

Total moles of reactants = \( 1 + 3 = 4 \) moles.

- On the **product side**:
- \( 2 NH_3 \) contributes 2 moles.

Total moles of products = \( 2 \) moles.

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

\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} = 2 - 4 = -2 \]

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

Now that we have \( \Delta n_g \), we can substitute it back into the equation for \( \Delta H \):

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

This simplifies to:

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

### Conclusion

Thus, we have expressed \( \Delta H \) in terms of \( \Delta U \) and \( RT \):

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