For the reaction, `2A (s) + 2B (g)to C (g) + D(s) at 298K`
`Delta U ^(@) =-10.5KJ and Delta S ^(@)=-44.10 JK^(-1)`
Calculate `Delta G ^(@)` for the reaction (in Joules).

Explanation

To calculate the standard change in Gibbs free energy (ΔG°) for the reaction given, we will follow these steps:

### Step 1: Identify the given data
We have the following information:
- ΔU° = -10.5 kJ
- ΔS° = -44.1 J/K
- Temperature (T) = 298 K

### Step 2: Convert ΔU° to Joules
Since we need ΔG° in Joules, we first convert ΔU° from kilojoules to joules:
\[ \Delta U° = -10.5 \text{ kJ} \times 1000 \text{ J/kJ} = -10500 \text{ J} \]

### Step 3: Calculate ΔH°
We need to find ΔH° using the relationship between ΔH° and ΔU°:
\[ \Delta H° = \Delta U° + \Delta N_g R T \]
where:
- ΔN_g = (moles of gaseous products - moles of gaseous reactants)
- R = 8.31 J/(mol·K)
- T = 298 K

### Step 4: Calculate ΔN_g
From the reaction \(2A (s) + 2B (g) \rightarrow C (g) + D (s)\):
- Moles of gaseous products = 1 (C)
- Moles of gaseous reactants = 2 (B)

Thus,
\[ \Delta N_g = 1 - 2 = -1 \]

### Step 5: Substitute values into the ΔH° equation
Now we can substitute the values into the ΔH° equation:
\[ \Delta H° = -10500 \text{ J} + (-1)(8.31 \text{ J/(mol·K)})(298 \text{ K}) \]
Calculating the second term:
\[ -1 \times 8.31 \times 298 = -2477.58 \text{ J} \]
Now substituting this back:
\[ \Delta H° = -10500 \text{ J} - 2477.58 \text{ J} = -12977.58 \text{ J} \]

### Step 6: Calculate ΔG°
Now we can calculate ΔG° using the formula:
\[ \Delta G° = \Delta H° - T \Delta S° \]
Substituting the values:
\[ \Delta G° = -12977.58 \text{ J} - (298 \text{ K})(-44.1 \text{ J/K}) \]
Calculating the second term:
\[ 298 \times -44.1 = -13173.8 \text{ J} \]
So,
\[ \Delta G° = -12977.58 \text{ J} + 13173.8 \text{ J} = 196.22 \text{ J} \]

### Final Answer
The standard change in Gibbs free energy (ΔG°) for the reaction is approximately:
\[ \Delta G° \approx 196.22 \text{ J} \]