Question

A certain reaction is spontaneous at ${72}^{\circ }\mathrm{C}$. If the enthalpy change for the reaction is $19\mathrm{kJ}/\mathrm{mol}$, what is the minimum value of $\mathrm{\Delta }S$ (in $\mathrm{J}/\mathrm{K}\cdot \mathrm{mol}$ ) for the reaction?

Short answer

The minimum $\mathrm{\Delta }S$ is approximately $55.04\text{J/K}\cdot \text{mol}$.

Step-by-step solution

Convert Temperature to Kelvin

To solve this problem, we first need to convert the given temperature from Celsius to Kelvin. The formula for conversion is $T(\text{K})=T{(}^{\circ }C)+273.15$. Substituting the given temperature: $T=72+273.15=345.15\text{K}$.

Use the Gibbs Free Energy Equation

Spontaneity of a reaction can be determined using the Gibbs Free Energy change equation: $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$. For a reaction to be spontaneous, $\mathrm{\Delta }G\le 0$.

Solve for $\mathrm{\Delta }S$

Since the reaction is spontaneous at 72°C, we set $\mathrm{\Delta }G\le 0$. Rearrange the equation $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$ to $T\mathrm{\Delta }S\ge \mathrm{\Delta }H$. Solving for $\mathrm{\Delta }S$, we have $\mathrm{\Delta }S\ge \frac{\mathrm{\Delta }H}{T}$.

Substitute Values into the Rearranged Equation

We substitute the given $\mathrm{\Delta }H=19\text{kJ/mol}=19000\text{J/mol}$ and $T=345.15\text{K}$ into the equation $\mathrm{\Delta }S\ge \frac{19000}{345.15}$.

Calculate $\mathrm{\Delta }S$

Perform the division: $\mathrm{\Delta }S\ge \frac{19000}{345.15}\approx 55.04\text{J/K}\cdot \text{mol}$. Thus, the minimum value of $\mathrm{\Delta }S$ is approximately $55.04\text{J/K}\cdot \text{mol}$.

Key concepts

Enthalpy Change

Enthalpy change, often represented as $\mathrm{\Delta }H$, measures the heat absorbed or released during a chemical reaction at constant pressure. It indicates whether a reaction is endothermic or exothermic.

- **Endothermic reactions** absorb heat from their surroundings, leading to a positive $\mathrm{\Delta }H$.
- **Exothermic reactions** release heat, resulting in a negative $\mathrm{\Delta }H$.

In the exercise, the given enthalpy change is 19 kJ/mol, indicating the amount of heat absorbed to make the reaction potentially spontaneous at 72°C. When considering spontaneity, the enthalpy change of 19 kJ/mol implies the reaction might need a specific entropy change value to occur spontaneously under these conditions. Converting enthalpy from kJ to J is crucial for consistency: 19 kJ/mol converts to 19000 J/mol. This step ensures that all quantities have compatible units when substituted into the Gibbs Free Energy equation.

Entropy Change

Entropy, indicated as $\mathrm{\Delta }S$, describes the disorder or randomness of a system. Greater entropy implies more disorderly states. For a reaction:

- An increase in entropy $(\mathrm{\Delta }S>0)$ typically indicates more disorder post-reaction.
- A decrease $(\mathrm{\Delta }S<0)$ signals a more ordered state.

Entropy plays a vital role in determining reaction spontaneity, alongside enthalpy. In the exercise, we calculate the minimum equilibrium value of $\mathrm{\Delta }S$ to ensure spontaneity at the given conditions. Using the relationship expressed in the Gibbs Free Energy equation $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$ and knowing that $\mathrm{\Delta }G\le 0$, we can rearrange to find $T\mathrm{\Delta }S\ge \mathrm{\Delta }H$. Thus, for a reaction at 345.15 K, $\mathrm{\Delta }S$ must satisfy $\mathrm{\Delta }S\ge \frac{19000}{345.15}\approx 55.04\text{J/K}\cdot \text{mol}$. This ensures sufficient disorder for the reaction to proceed spontaneously.

Spontaneity of Reactions

A spontaneous reaction is one that occurs naturally under specific conditions without external intervention. The Gibbs Free Energy change $\mathrm{\Delta }G$ is used to gauge this spontaneity:

- If $\mathrm{\Delta }G<0$, the reaction is spontaneous.
- If $\mathrm{\Delta }G=0$, the system is at equilibrium.
- If $\mathrm{\Delta }G>0$, the reaction is non-spontaneous.

In the context of the exercise, understanding spontaneity involves using the Gibbs equation: $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$. The given conditions make the reaction spontaneous at 72°C. Thus, it is essential that the sum of the enthalpy and entropy changes results in a negative $\mathrm{\Delta }G$. This is why calculating the correct $\mathrm{\Delta }S$ is necessary: it ensures that $\mathrm{\Delta }G$ remains non-positive, confirming that the reaction can spontaneously proceed at the specified temperature.