Gibbs Helmholtz equation relates the enthalpy, entropy and free energy change of the process at constant pressure and temperature as
`DeltaG=DeltaH-TDeltaS " (at constant P, T)"`
In General the magnitude of `DeltaH` does not change much with the change in temperature but the terms `TDeltaS` changes appreciably. Hence in some process spontaneity is very much dependent on temperature and such processes are generally known as entropy driven process.
Fro the reaction at 298 K, `A_(2)B_(4)rarr2AB_(2)`
`DeltaH=2" kJ"` and `DeltaS` = 20 J/K at constant P and T, the reaction will be

Options

A. spontaneous and entropy driven
B. spontaneous and enthalpy driven
C. non-spontaneous
D. at equilibrium

Explanation

To determine whether the reaction \( A_2B_4 \rightarrow 2AB_2 \) is spontaneous at 298 K, we can use the Gibbs-Helmholtz equation:

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

### Step 1: Identify the given values
- \(\Delta H = 2 \, \text{kJ} = 2000 \, \text{J}\) (since \(1 \, \text{kJ} = 1000 \, \text{J}\))
- \(\Delta S = 20 \, \text{J/K}\)
- \(T = 298 \, \text{K}\)

### Step 2: Substitute the values into the Gibbs-Helmholtz equation
Now, we substitute the values into the equation:

\[ \Delta G = 2000 \, \text{J} - (298 \, \text{K} \times 20 \, \text{J/K}) \]

### Step 3: Calculate \(T \Delta S\)
Calculate \(T \Delta S\):

\[ T \Delta S = 298 \, \text{K} \times 20 \, \text{J/K} = 5960 \, \text{J} \]

### Step 4: Calculate \(\Delta G\)
Now, substitute \(T \Delta S\) back into the equation for \(\Delta G\):

\[ \Delta G = 2000 \, \text{J} - 5960 \, \text{J} = -3960 \, \text{J} \]

### Step 5: Analyze the result
Since \(\Delta G = -3960 \, \text{J}\) is negative, the reaction is spontaneous.

### Step 6: Determine the nature of the process
The reaction is spontaneous due to the negative contribution from the \(T \Delta S\) term, indicating that it is an entropy-driven process.

### Final Answer
The reaction \( A_2B_4 \rightarrow 2AB_2 \) is spontaneous and is an entropy-driven process.

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