Question

Consider the following equilibrium constant versus temperature data for some reaction: $$\begin{array}{ll}\mathit{T}\left({}^{\circ }\mathbf{C}\right)& \text{ K }\\ 109& 2.54\times {10}^4\\ 225& 5.04\times {10}^2\\ 303& 6.33\times {10}^1\\ 412& 2.25\times {10}^{-1}\\ 539& 3.03\times {10}^{-3}\end{array}$$ Predict the signs for $\mathrm{\Delta }{G}^{\circ },\mathrm{\Delta }{H}^{\circ },$ and $\mathrm{\Delta }{S}^{\circ }$ for this reaction at ${25}^{\circ }\mathrm{C}$ . Assume $\mathrm{\Delta }{H}^{\circ }$ and $\mathrm{\Delta }{S}^{\circ }$ do not depend on temperature.

Short answer

In summary, for the given reaction at 25°C, we have the following signs: ∆G⁰ is negative, which indicates a spontaneous reaction. ∆H⁰ is negative, which indicates an exothermic reaction. ∆S° is negative, which indicates a decrease in the system's entropy.

Step-by-step solution

Convert the temperatures from Celsius to Kelvin

Since the van't Hoff equation uses temperature in Kelvin (K), we need to convert the given temperatures from Celsius to Kelvin. To do so, add 273.15 to each temperature in the table: $$T(K)=T{(}^{\circ }C)+273.15$$

Choose two temperature-equilibrium constant pairs from the table

We will pick two pairs of data points (T and K) from the table to apply the van't Hoff equation. For the sake of this solution, let's use the first and second data points: T₁ = 109 °C = 382.15 K and K₁ = 2.54 × 10⁴ T₂ = 225 °C = 498.15 K and K₂ = 5.04 × 10²

Apply the van't Hoff equation

The van't Hoff equation relates the temperature and equilibrium constant to the standard Gibbs free energy change (∆G°), standard enthalpy change (∆H°), and standard entropy change (∆S°): $$\frac{d\ln K}{dT}=\frac{\mathrm{\Delta }{H}^{\circ }}{R{T}^2}$$ Since ∆H° and ∆S° are assumed to be constant with respect to temperature, we can express the van't Hoff equation for our two data points: $$\frac{\ln K_2-\ln K_1}{T_2-T_1}=\frac{\mathrm{\Delta }{H}^{\circ }}{R}(\frac{1}{T_2}-\frac{1}{T_1})$$

Calculate ∆H⁰

Solve the van't Hoff equation for ∆H⁰ by substituting the selected data points into the equation: $$\mathrm{\Delta }{H}^{\circ }=R\frac{\ln K_2-\ln K_1}{\frac{1}{T_2}-\frac{1}{T_1}}$$ Where R is the gas constant, R = 8.314 J/(mol·K). $$\mathrm{\Delta }{H}^{\circ }=8.314\frac{\ln (5.04\times {10}^2)-\ln (2.54\times {10}^4)}{\frac{1}{498.15}-\frac{1}{382.15}}$$ After calculation, we get: $$\mathrm{\Delta }{H}^{\circ }=-32,230.53\text{J/mol}$$ Since ∆H⁰ is negative, the reaction is exothermic.

Calculate ∆G⁰

At 25°C (298.15 K), we will use the data point with the closest temperature to calculate ∆G⁰: T = 303 °C = 576.15 K and K = 6.33 × 10¹ Use the Gibbs free energy equation to calculate ∆G°: $$\mathrm{\Delta }{G}^{\circ }=-RT\ln K$$ $$\mathrm{\Delta }{G}^{\circ }=-8.314\times 298.15\times \ln (6.33\times {10}^1)$$ After calculation, we get: $$\mathrm{\Delta }{G}^{\circ }=-8,344.96\text{J/mol}$$ Since ∆G⁰ is negative, the reaction is spontaneous at 25°C.

Calculate ∆S⁰

Use the relationship between ∆G°, ∆H°, and ∆S°: $$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$$ Rearrange to solve for ∆S°: $$\mathrm{\Delta }{S}^{\circ }=\frac{\mathrm{\Delta }{H}^{\circ }-\mathrm{\Delta }{G}^{\circ }}{T}$$ $$\mathrm{\Delta }{S}^{\circ }=\frac{-32,230.53\text{J/mol}-(-8,344.96\text{J/mol})}{298.15\text{K}}$$ After calculation, we get: $$\mathrm{\Delta }{S}^{\circ }=-80.30\text{J/(mol·K)}$$ Since ∆S⁰ is negative, the reaction leads to a decrease in the system's entropy.

Conclusion

In summary, for the given reaction at 25°C, we have the following signs: ∆G⁰ is negative, which indicates a spontaneous reaction. ∆H⁰ is negative, which indicates an exothermic reaction. ∆S° is negative, which indicates a decrease in the system's entropy.

Key concepts

Gibbs free energy

Gibbs free energy, often denoted as $\mathrm{\Delta }{G}^{\circ }$, is a vital concept in thermodynamics that determines the spontaneity of a reaction at constant pressure and temperature.
It is calculated using the equation: $$\mathrm{\Delta }{G}^{\circ }=\mathrm{\Delta }{H}^{\circ }-T\mathrm{\Delta }{S}^{\circ }$$ where $\mathrm{\Delta }{H}^{\circ }$ is the change in enthalpy, $T$ is the temperature in Kelvin, and $\mathrm{\Delta }{S}^{\circ }$ is the change in entropy.
- If $\mathrm{\Delta }{G}^{\circ }<0$, the reaction is spontaneous under the given conditions, meaning it can proceed without the input of additional energy.- If $\mathrm{\Delta }{G}^{\circ }>0$, the reaction is non-spontaneous, indicating that it would require energy to proceed.In the exercise, the calculated $\mathrm{\Delta }{G}^{\circ }$ value was negative, indicating that the reaction proceeds spontaneously at 25°C. This means that under these conditions, the system has a natural tendency to proceed towards equilibrium.

enthalpy change

Enthalpy change, symbolized by $\mathrm{\Delta }{H}^{\circ }$, is a measure of the total heat content in a thermodynamic system.
It describes how much heat energy is absorbed or released during a chemical reaction at constant pressure.**Key Points about Enthalpy Change**:

• **Exothermic Reactions**: If $\mathrm{\Delta }{H}^{\circ }<0$, the reaction releases heat to its surroundings, making it exothermic.
• **Endothermic Reactions**: If $\mathrm{\Delta }{H}^{\circ }>0$, the reaction absorbs heat, making it endothermic.

In our provided exercise, the calculated $\mathrm{\Delta }{H}^{\circ }$ was negative, meaning the reaction is exothermic.
This indicates that energy is given off as heat when the reaction occurs, which is typical for reactions where bonds form, releasing energy.

entropy change

Entropy change, represented by $\mathrm{\Delta }{S}^{\circ }$, is a measure of the disorder or randomness in a thermodynamic system.
A system's entropy reflects how energy is distributed within it, often determining the feasibility of states or reactions.**Understanding Entropy Change**:

• Positive $\mathrm{\Delta }{S}^{\circ }$: The system's disorder increases. In chemical reactions, this might happen when gases are produced from liquids or solids, or when a mixture is created from separate substances.
• Negative $\mathrm{\Delta }{S}^{\circ }$: The system's disorder decreases. This frequently occurs when gases condense into liquids or solids, indicating energy focusing in order.

In the given exercise, the obtained $\mathrm{\Delta }{S}^{\circ }$ was negative, signifying a reduction in the system's entropy.
This usually implies a more ordered state due to the reaction process, reflecting a decrease in randomness as the products form.