The value of `DeltaH and DeltaS` for two reactions are given below :
Reaction A : `DeltaH = -10.5 xx 10^(3) J "mol"^(-1)`
`DeltaS = + 31 J K^(-1) "mol"^(-1)`
Reaction B : `DeltaH = -11.7 xx 10^(3) J "mol"^(-1)`
`DeltaS = -105 J K^(-1) "mol"^(-1)`
Decide whether these reactions are spontaneous or not at 298 K.

Explanation

To determine the spontaneity of the given reactions at 298 K, we will calculate the Gibbs free energy change (ΔG) for each reaction using the formula:

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

where:
- ΔG = Gibbs free energy change
- ΔH = Change in enthalpy
- T = Temperature in Kelvin
- ΔS = Change in entropy

### Step 1: Calculate ΔG for Reaction A

Given:
- ΔH (Reaction A) = -10.5 × 10³ J/mol
- ΔS (Reaction A) = +31 J/K·mol
- T = 298 K

Using the formula:

\[ \Delta G_A = \Delta H_A - T \Delta S_A \]

Substituting the values:

\[ \Delta G_A = (-10.5 \times 10^3) - (298 \times 31) \]

Calculating \(298 \times 31\):

\[ 298 \times 31 = 9228 \, \text{J/mol} \]

Now substituting back into the equation:

\[ \Delta G_A = -10,500 - 9228 = -1,9728 \, \text{J/mol} \]

### Step 2: Determine spontaneity for Reaction A

Since ΔG_A is negative:

\[ \Delta G_A < 0 \implies \text{Reaction A is spontaneous} \]

### Step 3: Calculate ΔG for Reaction B

Given:
- ΔH (Reaction B) = -11.7 × 10³ J/mol
- ΔS (Reaction B) = -105 J/K·mol
- T = 298 K

Using the formula:

\[ \Delta G_B = \Delta H_B - T \Delta S_B \]

Substituting the values:

\[ \Delta G_B = (-11.7 \times 10^3) - (298 \times -105) \]

Calculating \(298 \times -105\):

\[ 298 \times -105 = -31,290 \, \text{J/mol} \]

Now substituting back into the equation:

\[ \Delta G_B = -11,700 + 31,290 = 19,590 \, \text{J/mol} \]

### Step 4: Determine spontaneity for Reaction B

Since ΔG_B is positive:

\[ \Delta G_B > 0 \implies \text{Reaction B is non-spontaneous} \]

### Final Conclusion

- Reaction A is spontaneous at 298 K.
- Reaction B is non-spontaneous at 298 K.

---