To determine if a chemical reaction can occur spontaneously (is feasible) at a given temperature, we must account for both the enthalpy change (\( \Delta H \)) and the entropy change (\( \Delta S \)) of the system. The thermodynamic function that combines these two properties is the Gibbs free energy change.
🔑 Key Principle
A chemical reaction is thermodynamically feasible (spontaneous) at a given temperature if and only if the change in Gibbs free energy is negative or zero (\( \Delta G \le 0 \)).
The Gibbs Free Energy Equation
The thermodynamic property defined by the relationship:
\( \Delta G = \Delta H - T\Delta S \)
Where \( \Delta H \) is the enthalpy change in \( \text{J mol}^{-1} \), \( T \) is the temperature in Kelvin (\( \text{K} \)), and \( \Delta S \) is the entropy change in \( \text{J K}^{-1}\text{mol}^{-1} \).
A very common trap in exams is unit mismatch! Enthalpy values are usually given in \( \text{kJ mol}^{-1} \) and entropy values in \( \text{J K}^{-1}\text{mol}^{-1} \). Before calculating \( \Delta G \), you must convert your entropy change to \( \text{kJ K}^{-1}\text{mol}^{-1} \) by dividing by 1000. Alternatively, multiply \( \Delta H \) by 1000 to convert it to Joules.
Feasibility and Temperature
Because \( T \) is always positive in Kelvin, the sign of \( \Delta G \) depends on the signs of \( \Delta H \) and \( \Delta S \). We can identify four distinct cases:
| \( \Delta H \) | \( \Delta S \) | \( -T\Delta S \) | \( \Delta G \) | Feasibility |
|---|---|---|---|---|
| Negative (−) | Positive (+) | Negative (−) | Always Negative (−) | Feasible at all temperatures |
| Positive (+) | Negative (−) | Positive (+) | Always Positive (+) | Never feasible |
| Negative (−) | Negative (−) | Positive (+) | Negative at low T | Feasible only at low temperatures |
| Positive (+) | Positive (+) | Negative (−) | Negative at high T | Feasible only at high temperatures |
Calculating the Threshold Temperature
For reactions where \( \Delta H \) and \( \Delta S \) have the same sign, the feasibility changes at a specific temperature. The boundary between feasible and non-feasible is when \( \Delta G = 0 \).
To find this transition temperature (often called the threshold temperature), we set \( \Delta G = 0 \):
a) Calculate \( \Delta G^\theta \) for this reaction at \( 298\text{ K} \) and state whether the reaction is feasible.
b) Calculate the temperature at which the reaction just ceases to be feasible.
Part a): Convert \( \Delta S^\theta \) to \( \text{kJ K}^{-1}\text{mol}^{-1} \):
\( \Delta S^\theta = \frac{-198.3}{1000} = -0.1983\text{ kJ K}^{-1}\text{mol}^{-1} \)
Substitute into the Gibbs equation:
\( \Delta G^\theta = \Delta H^\theta - T\Delta S^\theta \)
\( \Delta G^\theta = -92.2 - (298 \times -0.1983) \)
\( \Delta G^\theta = -92.2 + 59.1 = -33.1\text{ kJ mol}^{-1} \)
Because \( \Delta G^\theta < 0 \), the reaction is feasible at \( 298\text{ K} \).
Part b): The reaction ceases to be feasible when \( \Delta G = 0 \):
\( T = \frac{\Delta H^\theta}{\Delta S^\theta} = \frac{-92.2}{-0.1983} = 465\text{ K} \)
Since both \( \Delta H \) and \( \Delta S \) are negative, the reaction is feasible at low temperatures. Therefore, it is feasible below \( 465\text{ K} \) and ceases to be feasible at temperatures above \( 465\text{ K} \).
Plotting Free Energy against Temperature
The Gibbs equation can be rearranged into the form of a straight-line equation (\( y = mx + c \)):
If we plot \( \Delta G \) (on the y-axis) against \( T \) (on the x-axis):
- The y-intercept represents the enthalpy change (\( \Delta H \)).
- The gradient of the line represents the negative of the entropy change (\( -\Delta S \)).
- The x-intercept (where \( \Delta G = 0 \)) represents the threshold temperature \( T \).
Limitations of Predicting Feasibility
A term used to describe a system where a reaction is thermodynamically feasible (\( \Delta G < 0 \)) but does not occur at a measurable rate because the activation energy is too high.
For example, the combustion of glucose has a highly negative free energy change:
\( \text{C}_6\text{H}_{12}\text{O}_6(s) + 6\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 6\text{H}_2\text{O}(l) \quad \Delta G^\theta = -2870\text{ kJ mol}^{-1} \)
However, a bowl of glucose (sugar) can sit open to the air at room temperature for years without burning. This is because the activation energy for the reaction is extremely high, meaning the rate of reaction is negligible at room temperature. We say the reactants are kinetically stable.
While a negative \( \Delta G \) indicates that the reaction is thermodynamically feasible, it does not provide any information about the rate of reaction (kinetics).
The reaction may have a very high activation energy (\( E_a \)). At room temperature, very few colliding reactant particles will possess energy equal to or greater than the activation energy, resulting in a negligible rate of reaction. A catalyst or an increase in temperature is required to allow the reaction to proceed.
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