Question

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 ____ $\text{K}$.

Answer 1

Correct Answer: 2000

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.

Answer 2

For spontaneity, we use the Gibbs free energy equation where \( \Delta G = 0 \) at the threshold temperature for spontaneity:

\[ T = \frac{\Delta H}{\Delta S} = \frac{400}{0.2} = 2000 \, \text{K} \]