QJEEM24Ch3774: JEE Main 2024 Chemistry - Class 11 JEEM24Ch3774 Solution: For the reaction at and . The reaction will become spontaneous…

To determine the temperature above which the reaction $2A+B \rightarrow C$ becomes spontaneous, we can use the Gibbs free energy equation:

$\Delta G = \Delta H - T\Delta S$

The reaction becomes spontaneous when $\Delta G$ is negative. Therefore, we need to find the temperature at which $\Delta G$ changes from positive to negative. We set $\Delta G$ to zero to find the threshold temperature:

$0 = \Delta H - T\Delta S$

Substituting the given values of $\Delta H = 400 \, \text{kJ mol}^{-1} = 400,000 \, \text{J mol}^{-1}$ (since 1 kJ = 1000 J) and $\Delta S = 0.2 \, \text{kJ mol}^{-1} K^{-1} = 200 \, \text{J mol}^{-1} K^{-1}$ , we get:

$0 = 400,000 \, \text{J mol}^{-1} - T(200 \, \text{J mol}^{-1} K^{-1})$

Solving for $T$ , we have:

$T = \frac{400,000 \, \text{J mol}^{-1}}{200 \, \text{J mol}^{-1} K^{-1}}$

$T = 2000 \, \text{K}$

Therefore, the reaction will become spontaneous above $2000 \, \text{K}$ . This means that at temperatures higher than 2000 K, the reaction tends towards product formation without the need for external energy to drive the process.