For reaction, `2K_((g))+L_((g))rarr2M_((g)),DeltaU^(@)=-"10.5 KJ and "DeltaS^(@)=-"44.1 J K"^(-1)`. Calculate `DeltaG^(@)` for the reaction and predict whether the reaction will be spontaneous or non-spontaneous?

Options

A. `DeltaG=+"0.16 kJ, non-spontaneous"`
B. `DeltaG=-"0.16 kJ, spontaneous"`
C. `DeltaG=+"26.12 kJ, non-spontaneous"`
D. `DeltaG=-"26.12 kJ, spontaneous"`

Explanation

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction given the standard internal energy change (ΔU°) and the standard entropy change (ΔS°).

### Step-by-Step Solution:

1. **Identify Given Values:**
- ΔU° = -10.5 kJ (standard internal energy change)
- ΔS° = -44.1 J/K (standard entropy change)
- Temperature (T) = 298 K (standard temperature)
- R = 8.314 × 10^(-3) kJ/(K·mol) (gas constant)

2. **Calculate ΔN_g (Change in Moles of Gases):**
- From the reaction:
\[ 2K(g) + L(g) \rightarrow 2M(g) \]
- Moles of products = 2 (from 2M)
- Moles of reactants = 2 (from 2K) + 1 (from L) = 3
- Therefore,
\[ ΔN_g = \text{moles of products} - \text{moles of reactants} = 2 - 3 = -1 \]

3. **Calculate ΔH° (Standard Enthalpy Change):**
- Using the formula:
\[ ΔH° = ΔU° + ΔN_g \cdot R \cdot T \]
- Substitute the values:
\[ ΔH° = -10.5 \text{ kJ} + (-1) \cdot (8.314 \times 10^{-3} \text{ kJ/(K·mol)}) \cdot (298 \text{ K}) \]
- Calculate:
\[ ΔH° = -10.5 \text{ kJ} - 2.477 \text{ kJ} \]
\[ ΔH° = -12.977 \text{ kJ} \]
- Round it to:
\[ ΔH° \approx -12.98 \text{ kJ} \]

4. **Calculate ΔG° (Standard Gibbs Free Energy Change):**
- Using the formula:
\[ ΔG° = ΔH° - T \cdot ΔS° \]
- Convert ΔS° from J/K to kJ/K:
\[ ΔS° = -44.1 \text{ J/K} = -0.0441 \text{ kJ/K} \]
- Substitute the values:
\[ ΔG° = -12.98 \text{ kJ} - (298 \text{ K} \cdot -0.0441 \text{ kJ/K}) \]
- Calculate:
\[ ΔG° = -12.98 \text{ kJ} + 13.14 \text{ kJ} \]
\[ ΔG° = 0.16 \text{ kJ} \]

5. **Determine Spontaneity:**
- Since ΔG° = +0.16 kJ (which is greater than 0), the reaction is non-spontaneous.

### Final Answer:
- ΔG° = +0.16 kJ
- The reaction is non-spontaneous.