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

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.