Question

A reaction has \(K=1.9 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\) and \(K=9.1 \times 10^{3}\) at \(227^{\circ} \mathrm{C}\) . Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and \(\Delta S^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short answer

At 25°C, the signs for ΔG°, ΔH°, and ΔS° are as follows: ΔG° is negative, indicating a spontaneous reaction. ΔH° is positive, indicating an endothermic reaction where energy is absorbed by the system. ΔS° is positive, indicating an increase in entropy of the system.

Step-by-step solution

Write down the Van't Hoff equation

The Van't Hoff equation is given by: \[\frac{d(\ln K)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] We will first use this equation to solve for ΔH°.

Calculate the value of ΔH°

In order to find ΔH°, we will take the derivative of ln(K) with respect to T. To do this, we can use the information of the two equilibrium constants K1 at T1 = 25°C and K2 at T2 = 227°C. First, convert the temperatures from degree Celsius to Kelvin: \(T_1 = 25 + 273.15 = 298.15 K\) \(T_2 = 227 + 273.15 = 500.15 K\) Now, we can write: \[\frac{d(\ln K)}{dT} = \frac{\ln(K_2) - \ln(K_1)}{T_2 - T_1}\] \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{K_2}{K_1})}{T_2 - T_1}\] Substitute the given values of K1, K2, T1 and T2: \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{9.1\times 10^{3}}{1.9 \times 10^{-14}})}{500.15 - 298.15} = 0.08379 K^{-1}\] Using the Van't Hoff equation, we can now find ΔH°: \[\Delta H^{\circ} = \frac{d(\ln K)}{dT} \cdot RT^2\] \[\Delta H^{\circ} = 0.08379 \times 8.314 \times (298.15)^2 = 65686 J/mol\]

Determine the sign of ΔH° and ΔS°

Now that we have the value of ΔH°, we can determine its sign. Since ΔH° is positive (65686 J/mol), this indicates that the reaction is endothermic. This means that energy is absorbed by the system. Now we can determine the sign of ΔS° by checking which direction the value of K moves as temperature increases. In this case, at 25°C, K is very small (1.9 × 10⁻¹⁴) practically zero. However, at 227°C, K has increased significantly (9.1 × 10³). Therefore, we can deduce that ΔS° is positive, meaning the entropy of the system increases.

Calculate ΔG° and predict its sign

Since we know the signs of ΔH° and ΔS°, we can determine the sign of ΔG° at 25°C using the Gibbs-Helmholtz equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] We can predict the sign of ΔG° without calculating it (although we do not have the value of ΔS°). Since ΔH° is positive and ΔS° is positive, at 25°C (298.15 K), the ΔG° will be negative because the TΔS° term will be larger than the ΔH° term. Therefore, the reaction is spontaneous at 25°C, and we can conclude that the signs for ΔG°, ΔH°, and ΔS° at 25°C are as follows: ΔG°: Negative ΔH°: Positive ΔS°: Positive

Key concepts

Gibbs Free Energy

Gibbs Free Energy, represented as \(\Delta G^{\circ}\), is an essential concept in thermodynamics that helps us understand whether a reaction is spontaneous. A reaction is considered spontaneous if it occurs naturally without external influence. The formula for Gibbs Free Energy, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), clearly shows the relationship between enthalpy \(\Delta H^{\circ}\), temperature \(T\), and entropy \(\Delta S^{\circ}\).

• \(\Delta G^{\circ}\) Negative: The reaction is spontaneous.
• \(\Delta G^{\circ}\) Positive: The reaction is non-spontaneous.
• \(\Delta G^{\circ}\) Zero: The system is at equilibrium.

When both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, as in endothermic reactions increasing disorder, \(T\Delta S^{\circ}\) can outweigh \(\Delta H^{\circ}\), making the reaction spontaneous. This is exactly what happens in our exercise, leading to a negative \(\Delta G^{\circ}\), indicating the reaction's spontaneity despite the positive \(\Delta H^{\circ}\).

Van't Hoff Equation

The Van't Hoff equation is a valuable tool for understanding how the equilibrium constant \(K\) of a reaction depends on temperature. It is expressed as \[\frac{d(\ln K)}{dT} = \frac{\Delta H^{\circ}}{RT^2}.\]This equation tells us that the change in the logarithm of the equilibrium constant with temperature gives insights about \(\Delta H^{\circ}\), the reaction's enthalpy change.

• K increases with temperature if the reaction is endothermic (\(\Delta H^{\circ}\) is positive).
• K decreases with temperature if the reaction is exothermic (\(\Delta H^{\circ}\) is negative).

By observing how \(K\) changes, we can deduce whether a reaction absorbs or releases heat. In the exercise, the significant increase in \(K\) with temperature implies a positive \(\Delta H^{\circ}\), confirming the reaction is endothermic.

Equilibrium Constant

The equilibrium constant, \(K\), is a vital concept indicating the extent to which a reaction proceeds at a certain temperature. A large \(K\) means the products are favored, while a small \(K\) means the reactants are favored.

• If \(K >> 1\), the reaction mostly yields products.
• If \(K << 1\), the reaction remains mostly in the reactants form.

In our exercise, \(K\) changes dramatically from \(1.9 \times 10^{-14}\) at 25°C to \(9.1 \times 10^3\) at 227°C. This increase indicates that the reaction shifts significantly towards the products as the temperature rises. This behavior helps us understand the reaction's dynamics and its spontaneity as temperature changes. Larger \(K\) at higher temperatures suggests less reactant and more product formation.

Endothermic Reactions

Endothermic reactions are processes where heat is absorbed from the surroundings, indicated by a positive \(\Delta H^{\circ}\). When a reaction is endothermic:

• The system absorbs energy.
• \(\Delta H^{\circ}\) is greater than zero.
• Heat is required for the reaction to proceed.

In the provided exercise, the positive \(\Delta H^{\circ}\) calculated indicates an endothermic process. This means that the reaction absorbs heat, making it more product-favorable at higher temperatures. As temperature increases, the equilibrium shifts towards the formation of products, correlating with the increase in \(K\). Understanding such reactions helps in predicting how a reaction behaves as conditions change and designing systems that rely on temperature-sensitive reactions.