For the reaction at $$298 \mathrm{~K}, 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}, \Delta \mathrm{H}=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$$ and $\Delta S=0.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. The reaction will become spontaneous above __________ $\mathrm{K}$.

<p>To determine the temperature above which the reaction $2A+B \rightarrow C$ becomes spontaneous, we can use the Gibbs free energy equation:</p> <p>$\Delta G = \Delta H - T\Delta S$</p> <p>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:</p> <p>$0 = \Delta H - T\Delta S$</p> <p>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:</p> <p>$0 = 400,000 \, \text{J mol}^{-1} - T(200 \, \text{J mol}^{-1} K^{-1})$</p> <p>Solving for $T$, we have:</p> <p>$T = \frac{400,000 \, \text{J mol}^{-1}}{200 \, \text{J mol}^{-1} K^{-1}}$</p> <p>$T = 2000 \, \text{K}$</p> <p>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.</p>