Dependence of Spontaneity on Temperature:
For a process to be spontaneous , at constant temperature and pressure , there must be decrease in free energy of the system in the direction of the process , i.e. `DeltaG_(P.T) lt 0. DeltaG_(P.T) =0` implies the equilibrium condition and `DeltaG_(P.T) gt 0` corresponds to non- spontaneity.
Gibbs- Helmholtz equation relates the free energy change to the enthalpy and entropy changes of the process as : `" "DeltaG_(P.T) = DeltaH-TDeltaS" ""..."(1)`
The magnitude of `DeltaH` does not change much with the change in temperature but the entropy factor `TDeltaS` change appreciably . Thus, spontaneity of a process depends very much on temperature.
For endothermic process, both `DeltaH` and `DeltaS` are positive . The energy factor, the first factor of equation, opposes the spontaneity whereas entorpy factor favours it. At low temperature the favourable factor `TDeltaS` will be small and may be less than `DeltaH, DeltaG` will have positive value indicated the nonspontaneity of the process. On raising temperature , the factor `TDeltaS` Increases appreciably and when it exceeds `DeltaH, DeltaG` would become negative and the process would be spontaneous .
For an expthermic process, both `DeltaH` and `DeltaS` would be negative . In this case the first factor of eq.1 favours the spontaneity whereas the second factor opposes it. At high temperature , when `T DeltaS gt DeltaH, DeltaG` will have positive value, showing thereby the non-spontaneity fo the process . However , on decreasing temperature , the factor ,`TDeltaS` decreases rapidly and when `TDeltaS lt DeltaH, DeltaG` becomes negative and the process occurs spontaneously. Thus , an exothermic process may be spontaneous at low temperature and non-spontaneous at high temperature.
For the reaction at `25^(@), X_(2)O_(4)(l) rarr 2XO_(2)(g)`
`DeltaH=2.1 Kcal` and `DeltaS = 20 cal K^(-1)`. The reaction would be

Options

A. Spontaneous
B. non-spontaneous
C. at equilibrium
D. Unpredictable

Explanation

To determine the spontaneity of the reaction \( \text{X}_2\text{O}_4(l) \rightarrow 2\text{XO}_2(g) \) at \( 25^\circ C \) using the Gibbs-Helmholtz equation, we will follow these steps:

### Step 1: Write the Gibbs-Helmholtz equation
The Gibbs-Helmholtz equation is given by:
\[ \Delta G_{(P,T)} = \Delta H - T \Delta S \tag{1} \]
where:
- \( \Delta G \) = change in Gibbs free energy
- \( \Delta H \) = change in enthalpy
- \( T \) = temperature in Kelvin
- \( \Delta S \) = change in entropy

### Step 2: Convert given values to consistent units
We are given:
- \( \Delta H = 2.1 \, \text{kcal} \)
- \( \Delta S = 20 \, \text{cal K}^{-1} \)

First, convert \( \Delta H \) from kilocalories to calories:
\[ \Delta H = 2.1 \, \text{kcal} \times 1000 \, \text{cal/kcal} = 2100 \, \text{cal} \]

### Step 3: Convert temperature to Kelvin
The temperature given is \( 25^\circ C \). To convert this to Kelvin:
\[ T = 25 + 273.15 = 298.15 \, \text{K} \approx 298 \, \text{K} \]

### Step 4: Substitute values into the Gibbs-Helmholtz equation
Now substitute \( \Delta H \), \( T \), and \( \Delta S \) into the equation:
\[ \Delta G = 2100 \, \text{cal} - (298 \, \text{K} \times 20 \, \text{cal K}^{-1}) \]

### Step 5: Calculate \( T \Delta S \)
Calculate \( T \Delta S \):
\[ T \Delta S = 298 \times 20 = 5960 \, \text{cal} \]

### Step 6: Calculate \( \Delta G \)
Now, substitute this value back into the equation for \( \Delta G \):
\[ \Delta G = 2100 - 5960 = -3860 \, \text{cal} \]

### Step 7: Determine spontaneity
Since \( \Delta G < 0 \), the process is spontaneous.

### Final Answer
The reaction \( \text{X}_2\text{O}_4(l) \rightarrow 2\text{XO}_2(g) \) is spontaneous at \( 25^\circ C \).

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