For a certain reaction, ΔG θ = -45 kJ/mol and ΔH θ = -90 kJ/mol at 0 °C. What is the minimum temperature at which the reaction will become spontaneous, assuming that ΔH θ and ΔS θ are independent of temperature?
546 K
Calculating Reaction Spontaneity Temperature
The spontaneity of a chemical reaction is determined by the change in Gibbs free energy (\(\Delta G\)). For a reaction to be spontaneous, the Gibbs free energy change must be negative (\(\Delta G < 0\)). The relationship between Gibbs free energy (\(\Delta G\)), enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and absolute temperature (\(T\)) is given by the Gibbs-Helmholtz equation:
\[\Delta G = \Delta H - T\Delta S\]
We are given the standard Gibbs free energy change (\(\Delta G^\theta\)) and the standard enthalpy change (\(\Delta H^\theta\)) at a specific temperature (\(0^\circ\text{C}\)). We are also told to assume that \(\Delta H^\theta\) and \(\Delta S^\theta\) are independent of temperature.
Given data:
- Temperature, \(T_1 = 0^\circ\text{C}\). To use the Gibbs equation, temperature must be in Kelvin. \(T_1 = 0 + 273.15\text{ K} = 273.15\text{ K}\). We can approximate it as 273 K as is common in these types of problems.
- Standard Gibbs free energy change at \(T_1\), \(\Delta G^\theta_{T_1} = -45\text{ kJ/mol}\).
- Standard enthalpy change, \(\Delta H^\theta = -90\text{ kJ/mol}\).
- \(\Delta H^\theta\) and \(\Delta S^\theta\) are constant with temperature.
First, let's use the Gibbs-Helmholtz equation at the given temperature (\(273\text{ K}\)) to find the standard entropy change (\(\Delta S^\theta\)).
\[\Delta G^\theta_{T_1} = \Delta H^\theta - T_1\Delta S^\theta\]
Substitute the given values:
\[-45\text{ kJ/mol} = -90\text{ kJ/mol} - (273\text{ K})\Delta S^\theta\]
Rearrange the equation to solve for \(\Delta S^\theta\):
\[(273\text{ K})\Delta S^\theta = -90\text{ kJ/mol} - (-45\text{ kJ/mol})\]
\[(273\text{ K})\Delta S^\theta = -90\text{ kJ/mol} + 45\text{ kJ/mol}\]
\[(273\text{ K})\Delta S^\theta = -45\text{ kJ/mol}\]
\[\Delta S^\theta = \frac{-45\text{ kJ/mol}}{273\text{ K}}\]
\[\Delta S^\theta = -\frac{45}{273}\text{ kJ/mol} \cdot \text{K}^{-1}\]
Now, we need to find the minimum temperature at which the reaction becomes spontaneous. A reaction transitions between spontaneous and non-spontaneous when \(\Delta G = 0\). Let \(T_{boundary}\) be the temperature at which \(\Delta G^\theta = 0\). Assuming \(\Delta H^\theta\) and \(\Delta S^\theta\) are constant:
\[0 = \Delta H^\theta - T_{boundary}\Delta S^\theta\]
Substitute the value of \(\Delta H^\theta\) and the calculated \(\Delta S^\theta\):
\[0 = -90\text{ kJ/mol} - T_{boundary}\left(-\frac{45}{273}\text{ kJ/mol} \cdot \text{K}^{-1}\right)\]
\[0 = -90 + T_{boundary}\left(\frac{45}{273}\right)\text{ kJ/mol}\]
Rearrange to solve for \(T_{boundary}\):
\[T_{boundary}\left(\frac{45}{273}\right) = 90\]
\[T_{boundary} = \frac{90 \times 273}{45}\text{ K}\]
\[T_{boundary} = 2 \times 273\text{ K}\]
\[T_{boundary} = 546\text{ K}\]
At \(T = 546\text{ K}\), \(\Delta G^\theta = 0\). For the reaction to be spontaneous, \(\Delta G^\theta\) must be less than 0 (\(\Delta G^\theta < 0\)). Let's examine the sign of \(\Delta G^\theta\) relative to this temperature.
We have \(\Delta H^\theta < 0\) and \(\Delta S^\theta < 0\). For this combination of signs, spontaneity (\(\Delta G < 0\)) is favored at low temperatures. The condition for spontaneity is \(\Delta G = \Delta H - T\Delta S < 0\).
\[-90 - T\left(-\frac{45}{273}\right) < 0\]
\[-90 + T\left(\frac{45}{273}\right) < 0\]
\[T\left(\frac{45}{273}\right) < 90\]
\[T < \frac{90 \times 273}{45}\]
\[T < 546\text{ K}\]
The reaction is spontaneous when the temperature is less than 546 K. The temperature at which \(\Delta G\) transitions from negative to positive (or vice versa) is 546 K. While the phrasing "minimum temperature at which the reaction will become spontaneous" might seem counterintuitive with \(\Delta H < 0\) and \(\Delta S < 0\) (which favors spontaneity at low T), the options indicate that 546 K is the intended boundary temperature.
Therefore, 546 K is the temperature where the reaction is at equilibrium (\(\Delta G = 0\)). Below this temperature, the reaction is spontaneous. Above this temperature, it is non-spontaneous.
Revision Table: Key Thermodynamics Concepts
| Term | Symbol | Definition | Relation to Spontaneity |
|---|---|---|---|
| Gibbs Free Energy | \(\Delta G\) | Maximum reversible work obtainable from a system at constant temperature and pressure. | \(\Delta G < 0\): Spontaneous \(\Delta G = 0\): Equilibrium \(\Delta G > 0\): Non-spontaneous |
| Enthalpy Change | \(\Delta H\) | Heat absorbed or released during a reaction at constant pressure. | \(\Delta H < 0\): Exothermic (favors spontaneity at low T if \(\Delta S < 0\)) \(\Delta H > 0\): Endothermic (disfavors spontaneity) |
| Entropy Change | \(\Delta S\) | Change in disorder or randomness of a system. | \(\Delta S > 0\): Increase in disorder (favors spontaneity) \(\Delta S < 0\): Decrease in disorder (disfavors spontaneity) |
Additional Information: Spontaneity and Temperature
The effect of temperature on spontaneity depends on the signs of \(\Delta H\) and \(\Delta S\).
- If \(\Delta H < 0\) and \(\Delta S > 0\): \(\Delta G = (\text{negative}) - T(\text{positive})\). \(\Delta G\) is always negative. Spontaneous at all temperatures.
- If \(\Delta H > 0\) and \(\Delta S < 0\): \(\Delta G = (\text{positive}) - T(\text{negative}) = (\text{positive}) + T(\text{positive})\). \(\Delta G\) is always positive. Non-spontaneous at all temperatures.
- If \(\Delta H < 0\) and \(\Delta S < 0\): \(\Delta G = (\text{negative}) - T(\text{negative}) = (\text{negative}) + T(\text{positive})\). Spontaneous at low temperatures (where \(|-90| > |T(-45/273)|\)). Non-spontaneous at high temperatures. The crossover temperature is when \(\Delta G = 0\).
- If \(\Delta H > 0\) and \(\Delta S > 0\): \(\Delta G = (\text{positive}) - T(\text{positive})\). Non-spontaneous at low temperatures. Spontaneous at high temperatures (where \(|T\Delta S| > |\Delta H|\)). The crossover temperature is when \(\Delta G = 0\).
In this problem, we have \(\Delta H < 0\) and \(\Delta S < 0\), which means the reaction is spontaneous at temperatures below the boundary temperature where \(\Delta G = 0\). That boundary temperature is 546 K.
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