Entropy is a state function and its value depends on two or three variable temperature (T), Pressure(P) and volume (V). Entropy change for an ideal gas having number of moles(n) can be determined by the following equation.
`DeltaS=2.303 "nC"_(v)"log"((T_(2))/(T_(1))) + 2.303 "nR log" ((V_(2))/(V_(1)))`
`DeltaS=2.303 "nC"_(P) "log"((T_(2))/(T_(1))) + 2.303 "nR log" ((P_(1))/(P_(2)))`
Since free energy change for a process or a chemical equation is a deciding factor of spontaneity , which can be obtained by using entropy change `(DeltaS)` according to the expression, `DeltaG = DeltaH -TDeltaS` at a temperature `T`
For a reaction `M_(2)O(s) rarr 2M(s) + (1)/(2) O_(2)(g) , DeltaH = 30 KJ//mol` and `DeltaS=0.07 KJ/mole "at" 1` atm. Calculate upto which temperature the reaction would not be spontaneous.

Options

A. `T gt 428.6` K
B. `T gt 300.8` K
C. `T lt 300.8` K
D. `T lt 428.6` K

Explanation

To determine the temperature up to which the reaction \( M_2O(s) \rightarrow 2M(s) + \frac{1}{2} O_2(g) \) would not be spontaneous, we will use the Gibbs free energy equation:

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

### Step 1: Understand the spontaneity condition
A reaction is spontaneous when \( \Delta G < 0 \). Therefore, for the reaction to be non-spontaneous, we need:

\[ \Delta G \geq 0 \]

This leads us to the condition:

\[ \Delta H - T \Delta S \geq 0 \]

### Step 2: Rearranging the equation
Rearranging the equation gives us:

\[ \Delta H \geq T \Delta S \]

From this, we can express the temperature \( T \):

\[ T \leq \frac{\Delta H}{\Delta S} \]

### Step 3: Substitute the given values
We know from the problem:

- \( \Delta H = 30 \, \text{kJ/mol} = 30,000 \, \text{J/mol} \) (since we need to convert kJ to J)
- \( \Delta S = 0.07 \, \text{kJ/mol} = 70 \, \text{J/mol} \)

Now we substitute these values into the equation:

\[ T \leq \frac{30,000 \, \text{J/mol}}{70 \, \text{J/mol}} \]

### Step 4: Calculate the temperature
Calculating the right-hand side:

\[ T \leq \frac{30,000}{70} \approx 428.57 \, \text{K} \]

### Step 5: Conclusion
Thus, the reaction would not be spontaneous at temperatures greater than approximately \( 428.57 \, \text{K} \). Therefore, the temperature up to which the reaction would not be spontaneous is:

\[ T < 428.57 \, \text{K} \]