For the reaction `2NO_2 (g) hArr N_2O_4(g)`, when `DeltaS = - 176.0 J K^(-1) and DeltaH = - 57. 8 kJ mol^(-1)`, the magnitude of `Delta G` at 298 K for the reaction is __________ kJ `mol^(-1)`. (Nearest integer)

Answer

To solve the problem, we will use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Convert units of ΔS
Given:
- \(\Delta S = -176.0 \, \text{J K}^{-1}\)

We need to convert this to kJ K\(^{-1}\):
\[ \Delta S = -176.0 \, \text{J K}^{-1} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -0.176 \, \text{kJ K}^{-1} \]

### Step 2: Write down the values
Now we have:
- \(\Delta H = -57.8 \, \text{kJ mol}^{-1}\)
- \(T = 298 \, \text{K}\)
- \(\Delta S = -0.176 \, \text{kJ K}^{-1}\)

### Step 3: Substitute values into the Gibbs free energy equation
Now we substitute these values into the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
\[ \Delta G = -57.8 \, \text{kJ mol}^{-1} - (298 \, \text{K} \times -0.176 \, \text{kJ K}^{-1}) \]

### Step 4: Calculate \(T \Delta S\)
Calculating \(T \Delta S\):
\[ T \Delta S = 298 \times -0.176 = -52.448 \, \text{kJ mol}^{-1} \]

### Step 5: Substitute back to find ΔG
Now substituting back:
\[ \Delta G = -57.8 + 52.448 \]
\[ \Delta G = -5.352 \, \text{kJ mol}^{-1} \]

### Step 6: Find the magnitude of ΔG
The magnitude of \(\Delta G\) is:
\[ |\Delta G| = 5.352 \, \text{kJ mol}^{-1} \]

### Step 7: Round to the nearest integer
Rounding \(5.352\) to the nearest integer gives:
\[ |\Delta G| \approx 5 \, \text{kJ mol}^{-1} \]

### Final Answer
The magnitude of \(\Delta G\) at 298 K for the reaction is **5 kJ mol\(^{-1}\)**.

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