Consider the reaction, `N_(2)[g] + 3H_(2)[g] rarr 2NH_(3)[g]` carried out at constant temperature and pressure. If `Delta H and Delta U` are the enthalpy and internal energy changes for the reaction, which of the following expressions is true?

Options

A. `DeltaH gt Delta U`
B. `DeltaH lt Delta U`
C. `DeltaH = Delta U`
D. `DeltaH = 0`

Explanation

To solve the problem regarding the relationship between the enthalpy change (ΔH) and the internal energy change (ΔU) for the reaction \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), we can follow these steps:

### Step 1: Understand the relationship between ΔH and ΔU
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔU) for a reaction at constant temperature and pressure is given by the equation:

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

where:
- \( \Delta n_g \) = change in the number of moles of gas (moles of products - moles of reactants)
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Calculate Δng for the reaction
For the given reaction:

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

- **Moles of gaseous products**: 2 moles of \( NH_3 \)
- **Moles of gaseous reactants**: 1 mole of \( N_2 \) + 3 moles of \( H_2 \) = 4 moles

Now, calculate \( \Delta n_g \):

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

### Step 3: Substitute Δng into the ΔH equation
Now substitute \( \Delta n_g \) into the equation for ΔH:

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

This simplifies to:

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

### Step 4: Analyze the relationship
From the equation \( \Delta H = \Delta U - 2RT \), we can see that:

\[ \Delta H < \Delta U \]

This indicates that the change in enthalpy (ΔH) is less than the change in internal energy (ΔU) for the reaction.

### Conclusion
Thus, the correct expression is:

\[ \Delta U > \Delta H \]