At what temperature the following process would not be spontanous ?
`4Fe(s)+3O_(2)(g)to2Fe_(2)O_(3)(s)`
`DeltaH=-1648 kJ//"mole"`
`DeltaS=-560 J//"mole"`

Options

A. 2000 K
B. 3000 K
C. 2900 K
D. 2500 K

Answer

To determine the temperature at which the process `4Fe(s) + 3O2(g) → 2Fe2O3(s)` is not spontaneous, we can use the Gibbs free energy equation:

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

A process is non-spontaneous when \(\Delta G > 0\). Therefore, we need to find the temperature at which:

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

This can be rearranged to:

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

### Step 1: Convert \(\Delta H\) to the same units as \(\Delta S\)

Given:
- \(\Delta H = -1648 \, \text{kJ/mol}\)
- \(\Delta S = -560 \, \text{J/mol}\)

First, we need to convert \(\Delta H\) from kJ to J:

\[ \Delta H = -1648 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -1648000 \, \text{J/mol} \]

### Step 2: Substitute the values into the equation

Now we can substitute \(\Delta H\) and \(\Delta S\) into the equation:

\[ T < \frac{-1648000 \, \text{J/mol}}{-560 \, \text{J/mol}} \]

### Step 3: Calculate the temperature

Now, we perform the division:

\[ T < \frac{1648000}{560} \]

Calculating this gives:

\[ T < 2942.857 \, \text{K} \]

### Step 4: Conclusion

Thus, the temperature at which the process is not spontaneous is:

\[ T < 2943 \, \text{K} \]