The enthalpy change `(Delta H)` for the reaction
`N_(2) (g) + 3 H_(2) (g) rarr 2 NH_(3) (g)`
is `- 92.38 kJ` at `298 K`. The internal energy change `Delta U` at `298 K` is

Options

A. `- 92.38 kJ mol^(-)`
B. `- 87.42 kJ`
C. `- 97.34 kJ`
D. `- 89.9 kJ`

Explanation

To find the internal energy change (ΔU) for the reaction:

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

given that the enthalpy change (ΔH) is \(-92.38 \, \text{kJ}\) at \(298 \, \text{K}\), we can use the relationship between ΔH and ΔU, which 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 (products - reactants)
- \(R\) = universal gas constant (\(8.314 \, \text{J/mol·K}\) or \(0.008314 \, \text{kJ/mol·K}\))
- \(T\) = temperature in Kelvin

### Step 1: Calculate Δng

First, we need to calculate \(\Delta n_g\):

- Moles of gaseous products (NH₃): 2
- Moles of gaseous reactants (N₂ + 3H₂): 1 + 3 = 4

Now, calculate \(\Delta n_g\):

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

### Step 2: Substitute values into the equation

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

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

Substituting the known values:

\[ -92.38 \, \text{kJ} = \Delta U + (-2) \times (0.008314 \, \text{kJ/mol·K}) \times (298 \, \text{K}) \]

### Step 3: Calculate \(\Delta n_g RT\)

Now, calculate \(\Delta n_g RT\):

\[ \Delta n_g RT = -2 \times 0.008314 \times 298 \]

Calculating this gives:

\[ \Delta n_g RT = -2 \times 0.008314 \times 298 \approx -4.96 \, \text{kJ} \]

### Step 4: Solve for ΔU

Now, substitute this value back into the equation:

\[ -92.38 \, \text{kJ} = \Delta U - 4.96 \, \text{kJ} \]

Rearranging gives:

\[ \Delta U = -92.38 \, \text{kJ} + 4.96 \, \text{kJ} \]

Calculating this gives:

\[ \Delta U \approx -87.42 \, \text{kJ} \]

### Final Answer

Thus, the internal energy change (ΔU) at \(298 \, \text{K}\) is:

\[ \Delta U \approx -87.42 \, \text{kJ} \]