For the reaction,
`2A(g)+B(g) rarr 2D(g)`
`DeltaU^(Θ)=-10.5 kJ` and `DeltaS^(Θ)=-44.1 JK^(-1)`
Calculate `DeltaG^(Θ)` for the reaction, and predict whether the reaction may occur spontaneously.

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction and determine whether it is spontaneous. The reaction given is:

\[ 2A(g) + B(g) \rightarrow 2D(g) \]

### Step-by-Step Solution:

1. **Identify Given Data:**
- ΔU° = -10.5 kJ
- ΔS° = -44.1 J/K

2. **Convert ΔS° to kJ:**
Since ΔS° is given in J/K, we need to convert it to kJ/K for consistency with ΔU°.
\[ ΔS° = -44.1 \, \text{J/K} = -0.0441 \, \text{kJ/K} \]

3. **Calculate Δn (Change in Moles of Gas):**
- Products: 2 moles of D
- Reactants: 2 moles of A + 1 mole of B = 3 moles
\[ Δn = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]

4. **Calculate ΔH°:**
We use the relation:
\[ ΔH° = ΔU° + Δn \cdot R \cdot T \]
Where R = 8.314 J/(mol·K) = 0.008314 kJ/(mol·K) and T = 298 K (standard temperature).
\[ ΔH° = -10.5 \, \text{kJ} + (-1) \cdot (0.008314 \, \text{kJ/(mol·K)}) \cdot (298 \, \text{K}) \]
\[ ΔH° = -10.5 \, \text{kJ} - 2.478 \, \text{kJ} = -12.978 \, \text{kJ} \]

5. **Calculate ΔG°:**
Using the Gibbs free energy equation:
\[ ΔG° = ΔH° - T \cdot ΔS° \]
\[ ΔG° = -12.978 \, \text{kJ} - (298 \, \text{K} \cdot -0.0441 \, \text{kJ/K}) \]
\[ ΔG° = -12.978 \, \text{kJ} + 13.138 \, \text{kJ} = 0.160 \, \text{kJ} \]

6. **Determine Spontaneity:**
The reaction is spontaneous if ΔG° < 0. Since ΔG° = 0.160 kJ (which is positive), the reaction is non-spontaneous.

### Final Answer:
- ΔG° = 0.160 kJ
- The reaction is non-spontaneous.