For a chemical reaction A+ B `hArr C + D (Delta_r H^0 = 80 ` KJ `"mol"^(-1))` the entropy change `Delta_r S^0` depends on the temperature T (in K) as `(Delta_r S^0 = 2T(J K^(-1) "mol"^(-1))` .
Minimum temperature at which it will become spontaneous is ___________ K. (Integer)

Answer

To determine the minimum temperature at which the reaction becomes spontaneous, we can use the Gibbs free energy equation:

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

For the reaction to be spontaneous, \(\Delta G\) must be less than 0:

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

This can be rearranged to:

\[ \Delta H < T \Delta S \]

### Step 1: Convert \(\Delta H\) from kJ to J
Given that \(\Delta H = 80 \, \text{kJ/mol}\), we need to convert this to joules:

\[ \Delta H = 80 \times 10^3 \, \text{J/mol} = 80000 \, \text{J/mol} \]

### Step 2: Write the expression for \(\Delta S\)
The entropy change \(\Delta S\) is given as:

\[ \Delta S = 2T \, \text{(J K}^{-1} \text{mol}^{-1}) \]

### Step 3: Substitute \(\Delta H\) and \(\Delta S\) into the inequality
Now we substitute \(\Delta H\) and \(\Delta S\) into the inequality:

\[ 80000 < T \cdot (2T) \]

This simplifies to:

\[ 80000 < 2T^2 \]

### Step 4: Solve for \(T\)
Rearranging gives:

\[ T^2 > \frac{80000}{2} \]

\[ T^2 > 40000 \]

Taking the square root of both sides:

\[ T > \sqrt{40000} \]

Calculating the square root:

\[ T > 200 \, \text{K} \]

### Conclusion
The minimum temperature at which the reaction will become spontaneous is:

\[ \boxed{200} \, \text{K} \]