For the reaction
`N_(2)(g) + 3H_(2)(g) hArr 2NH_(3)(g), DeltaH=?`

Options

A. (a) `DeltaE+2RT`
B. (b) `DeltaE-2RT`
C. (c) `DeltaH=RT`
D. (d) `DeltaE-RT`

Explanation

To find the enthalpy change (ΔH) for the reaction:

\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \]

we can use the relationship between enthalpy change (ΔH), internal energy change (ΔE), and the change in the number of gaseous moles (ΔNG). The formula we will use is:

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

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

### Step-by-step Solution:

**Step 1: Identify the number of gaseous moles of products and reactants.**

- Products:
- 2 moles of \( NH_3(g) \)

- Reactants:
- 1 mole of \( N_2(g) \) + 3 moles of \( H_2(g) \) = 4 moles of gas

**Step 2: Calculate ΔNG.**

\[ \Delta N_G = \text{(moles of products)} - \text{(moles of reactants)} \]

\[ \Delta N_G = 2 - 4 = -2 \]

**Step 3: Substitute ΔNG into the equation for ΔH.**

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

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

This simplifies to:

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

**Step 4: Identify the correct option.**

From the derived equation, we see that:

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

Thus, the correct answer is option B: \( \Delta H = \Delta E - 2RT \).