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

Question

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

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The problem asks for the temperature above which the given reaction becomes spontaneous. We are provided with the standard enthalpy change ($\Delta H$) and standard entropy change ($\Delta S$) for the reaction at 298 K. Concept Used: The spontaneity of a reaction is determined by the sign of the Gibbs free energy change ($\Delta G$). The relationship between $\Delta G$, enthalpy change ($\Delta H$), and entropy change ($\Delta S$) at a constant temperature $T$ is given by the Gibbs-Helmholtz equation: $$ \Delta G = \Delta H - T\Delta S $$ For a reaction to be spontaneous, the value of $\Delta G$ must be negative ($\Delta G < 0$). The transition from a non-spontaneous to a spontaneous process occurs at the temperature where the system is at equilibrium, which is defined by the condition $\Delta G = 0$. This temperature is known as the equilibrium temperature, $T_{eq}$. Step-by-Step Solution: Step 1: Write down the given thermodynamic values for the reaction. $$ \Delta H = 400 \, \text{kJ mol}^{-1} $$ $$ \Delta S = 0.2 \, \text{kJ K}^{-1} \text{mol}^{-1} $$ Note that the units are consistent (both are in kJ). Step 2: Determine the condition for the reaction to become spontaneous. The reaction becomes spontaneous when $\Delta G < 0$. $$ \Delta H - T\Delta S < 0 $$ The threshold temperature for this change is the equilibrium temperature ($T_{eq}$), where $\Delta G = 0$. Step 3: Set $\Delta G = 0$ in the Gibbs-Helmholtz equation to find the equilibrium temperature. $$ 0 = \Delta H - T_{eq}\Delta S $$ Rearranging the equation to solve for $T_{eq}$: $$ T_{eq}\Delta S = \Delta H $$ $$ T_{eq} = \frac{\Delta H}{\Delta S} $$ Step 4: Substitute the given values of $\Delta H$ and $\Delta S$ into the equation to calculate $T_{eq}$. $$ T_{eq} = \frac{400 \, \text{kJ mol}^{-1}}{0.2 \, \text{kJ K}^{-1} \text{mol}^{-1}} $$ Final Computation & Result: Performing the calculation for the equilibrium temperature: $$ T_{eq} = 2000 \, \text{K} $$ This is the temperature at which the reaction is at equilibrium. Since both $\Delta H$ (enthalpy) and $\Delta S$ (entropy) are positive, the reaction is endothermic and becomes more disordered. For the reaction to be spontaneous ($\Delta G < 0$), the entropy term ($T\Delta S$) must be larger than the enthalpy term ($\Delta H$). This occurs at temperatures above the equilibrium temperature. Therefore, the condition for spontaneity is $T > T_{eq}$. The reaction will become spontaneous above 2000 K .

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

Correct Answer: 2000

Approach Solution - 1

Concept Used:

Step-by-Step Solution:

Final Computation & Result:

Approach Solution -2

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