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.
A reaction has a value of `DeltaH =-40 Kcal` at 400 k cal `mol^(-1)`. The reaction is spontaneous, below this temperature , it is not . The values fo `DeltaG` and `DeltaS` at 400 k are respectively

Options

A. `0,-0.1 cal K^(-1)`
B. `0,100 cal K^(-1)`
C. `-10` Kcal, `-100"cal"K^(-1)`
D. `0,-100 cal K^(-1)`

Explanation

To solve the problem regarding the dependence of spontaneity on temperature using the Gibbs-Helmholtz equation, we will follow these steps:

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

### Step 2: Analyze the Given Information
We are given:
- \(\Delta H = -40 \, \text{kcal}\) (which is \(-40,000 \, \text{cal}\)),
- Temperature \(T = 400 \, \text{K}\),
- The reaction is spontaneous below this temperature and non-spontaneous above it, indicating that at \(400 \, \text{K}\), the system is at equilibrium.

### Step 3: Set Up the Equation at Equilibrium
At equilibrium, \(\Delta G = 0\). Therefore, we can set up the equation:
\[ 0 = \Delta H - T \Delta S \]
Rearranging gives:
\[ \Delta H = T \Delta S \]

### Step 4: Calculate \(\Delta S\)
Substituting the known values:
\[ \Delta S = \frac{\Delta H}{T} = \frac{-40,000 \, \text{cal}}{400 \, \text{K}} \]
Calculating this gives:
\[ \Delta S = -100 \, \text{cal/K} \]

### Step 5: Determine \(\Delta G\) and \(\Delta S\)
Since we have established that at \(400 \, \text{K}\):
- \(\Delta G = 0\)
- \(\Delta S = -100 \, \text{cal/K}\)

### Final Answer
The values of \(\Delta G\) and \(\Delta S\) at \(400 \, \text{K}\) are:
- \(\Delta G = 0\)
- \(\Delta S = -100 \, \text{cal/K}\)