A process has `Delta H= 200` J `mol^(-1)` and `DeltaS=40" JK"^(-1)mol^(-1)`. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :

Options

A. 5K
B. 12 K
C. 4 K
D. 20 K

Explanation

To determine the minimum temperature above which the process will be spontaneous, we can use the Gibbs free energy equation:

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

For the process to be spontaneous, \(\Delta G\) must be less than or equal to zero:

\[ \Delta G < 0 \implies \Delta H - T \Delta S < 0 \]

Rearranging this inequality gives us:

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

From this, we can derive the minimum temperature \(T\) at which the process becomes spontaneous:

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

### Step 1: Identify the values of \(\Delta H\) and \(\Delta S\)

Given:
- \(\Delta H = 200 \, \text{J/mol}\)
- \(\Delta S = 40 \, \text{J/K/mol}\)

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

Now, substituting the values into the equation:

\[ T > \frac{200 \, \text{J/mol}}{40 \, \text{J/K/mol}} \]

### Step 3: Calculate the temperature

Calculating the right side:

\[ T > \frac{200}{40} = 5 \, \text{K} \]

### Conclusion

Therefore, the minimum temperature above which the process will be spontaneous is:

\[ T = 5 \, \text{K} \]

### Final Answer

The minimum temperature above which the process will be spontaneous is **5 K**.

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